NCERT Exemplar for Class 11 Maths Chapter 13 - Limits and Derivatives (Book Solutions)

Examples

Short Answer Type Questions

Example 1: Evaluate: $\mathop {\lim }\limits_{x \to 2} \dfrac{1}{{x - 2}} - \dfrac{{2(2x - 3)}}{{{x^3} - 3{x^2} + 2x}}$

Ans: Given: $\mathop {\lim }\limits_{x \to 2} \dfrac{1}{{x - 2}} - \dfrac{{2(2x - 3)}}{{{x^3} - 3{x^2} + 2x}}$

Let $f$ and $g$ be two functions such that both $\mathop {\lim }\limits_{x \to a} f(x)$ and $\mathop {\lim }\limits_{x \to a} g(x)$ exist. Then $\mathop {\lim }\limits_{x \to a} [f(x) - g(x)] = \mathop {\lim }\limits_{x \to a} f(x) - \mathop {\lim }\limits_{x \to a} g(x)$

$\mathop {\lim }\limits_{x \to 2} \dfrac{1}{{x - 2}} - \dfrac{{2(2x - 3)}}{{{x^3} - 3{x^2} + 2x}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{1}{{x - 2}} - \dfrac{{2(2x - 3)}}{{x(x - 1)(x - 2}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{x(x - 1) - 2(2x - 3)}}{{x(x - 1)(x - 2)}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 5x + 6}}{{x(x - 1)(x - 2)}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - 2)(x - 3)}}{{x(x - 1)(x - 2)}}\quad [x - 2 \ne 0]$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 3}}{{x(x - 1)}} = \dfrac{{ - 1}}{2}$ 

Example 2: Evaluate  $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2 + x}  - \sqrt 2 }}{x}$

Ans: Given: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2 + x}  - \sqrt 2 }}{x}$

Let $f$ and $g$ be two functions such that both $\mathop {\lim }\limits_{x \to a} f(x)$ and $\mathop {\lim }\limits_{x \to a} g(x)$ exist. Then $\mathop {\lim }\limits_{x \to a} [f(x) - g(x)] = \mathop {\lim }\limits_{x \to a} f(x) - \mathop {\lim }\limits_{x \to a} g(x)$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2 + x}  - \sqrt 2 }}{x}$

 Putting $y = 2 + x$ so that when$x \to 0,y \to 2$. 

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2 + x}  - \sqrt 2 }}{x}$

$= \mathop {\lim }\limits_{y \to 2} \dfrac{{{y^{\dfrac{1}{2}}} - {2^{\dfrac{1}{2}}}}}{{y - 2}}$

$= \dfrac{1}{2}{(2)^{\dfrac{1}{2} - 1}}$

$= \dfrac{1}{2} \cdot {2^{ - \dfrac{1}{2}}} = \dfrac{1}{{2\sqrt 2 }}$ 

Example 3: Find the positive integer $n$ so that  $\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^n} - {3^n}}}{{x - 3}} = 108$.

Ans: Given: $\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^n} - {3^n}}}{{x - 3}} = 108$

$\mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$ is to be used to solve the question.

$\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^n} - {3^n}}}{{x - 3}} = n{(3)^{n - 1}}$

$= n{(3)^{n - 1}}$

$= 108$

$= 4(27) = 4{(3)^{4 - 1}}$ 

Comparing with $\mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$

$n = 4$

$\therefore$ Positive integer $n$ is $4$

Example 4: Evaluate $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} (\sec x - \tan x)$.

Ans: Given: $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} (\sec x - \tan x)$

 Putting $y = \dfrac{\pi }{2} - x.$ Then $y \to 0$ as $x \to \dfrac{\pi }{2}.$

$= \mathop {\lim }\limits_{y \to 0} \left[ {\sec \left( {\dfrac{\pi }{2} - y} \right) - \tan \left( {\dfrac{\pi }{2} - y} \right)} \right]$

$= \mathop {\lim }\limits_{y \to 0} (\operatorname{cosec} y - \cot y)$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{1}{{\sin y}} - \dfrac{{\cos y}}{{\sin y}}$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{1 - \cos y}}{{\sin y}}$ 

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{2{{\sin }^2}\dfrac{y}{2}}}{{2\sin \dfrac{y}{2}\cos \dfrac{y}{2}}}\quad {\text{ }}\left( {\because {{\sin }^2}\dfrac{y}{2} = \dfrac{{1 - \cos y}}{2},\sin y = 2\sin \dfrac{y}{2}\cos \dfrac{y}{2}} \right)$

$= \mathop {\lim }\limits_{\dfrac{y}{2} \to 0} \tan \dfrac{y}{2} = 0$ 

Example 5: Evaluate $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin (2 + x) - \sin (2 - x)}}{x}$.

Ans: Given: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin (2 + x) - \sin (2 - x)}}{x}$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin (2 + x) - \sin (2 - x)}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos \dfrac{{(2 + x + 2 - x)}}{2}\sin \dfrac{{(2 + x - 2 + x)}}{2}}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos 2\sin x}}{x}$

$= 2\cos 2\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 2\cos 2\quad {\text{ }}\left( {{\text{as }}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right)$ 

Example 6: Find the derivative of $f(x) = ax + b$, where $a$ and $b$ are non-zero constants, by first principle. 

Ans: Given:$a$ and $b$ are non-zero constants

Suppose $f$ is a real valued function, then ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

$\because {f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{a(x + h) + b - (ax + b)}}{h}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{ax + ah + b - ax - b}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{bh}}{h} = b$ 

Example 7: Find the derivative of $f(x) = a{x^2} + bx + c$, where $a,b$ and $c$ are none-zero constant, by first principle.

Ans: Given: $a,b$ and $c$ are none-zero constant

Suppose $f$ is a real valued function, then ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{a{{(x + h)}^2} + b(x + h) + c - a{x^2} - bx - c}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{bh + a{h^2} + 2axh}}{h}$

$= \mathop {\lim }\limits_{h \to 0} ah + 2ax + b = b + 2ax$ 

Example 8: Find the derivative of $f(x) = {x^3}$, by first principle. 

Ans: Given: $f(x) = {x^3}$

Suppose $f$ is a real valued function, then ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

  ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

   $= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(x + h)}^3} - {x^3}}}{h}$

   $= \mathop {\lim }\limits_{h \to 0} \dfrac{{{x^3} + {h^3} + 3xh(x + h) - {x^3}}}{h}$

   $= \mathop {\lim }\limits_{h \to 0} \left( {{h^2} + 3x(x + h)} \right) = 3{x^2}$ 

Example 9: Find the derivative of $f(x) = \dfrac{1}{x}$ by first principle.

Ans: Given: $f(x) = \dfrac{1}{x}$

Suppose $f$ is a real valued function, then ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\dfrac{1}{{x + h}} - \dfrac{1}{x}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{ - h}}{{h(x + h)x}} = \dfrac{{ - 1}}{{{x^2}}}$ 

Example 10: Find the derivative of $f(x) = sinx$, by first principle. 

Ans: Given:$f(x) = \sin x$

${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (x + h) - \sin x}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{2\cos \dfrac{{2x + h}}{2}\sin \dfrac{h}{2}}}{{2 \cdot \dfrac{h}{2}}}$

$= \mathop {\lim }\limits_{h \to 0} \cos \dfrac{{(2x + h)}}{2} \cdot \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \dfrac{h}{2}}}{{\dfrac{h}{2}}}$

$= \cos x.1 = \cos x$

Example 11: Find the derivative of $f(x) = {x^n}$, where $n$ is positive integer, by first principle.

Ans: Given:$f(x) = {x^n}$ where $n$ is positive integer

${f^\prime }(x) = \dfrac{{f(x + h) - f(x)}}{h}$

$= \dfrac{{{{(x + h)}^n} - {x^n}}}{h}$ 

Using Binomial theorem, ${(x + h)^n}{ = ^n}{{\text{C}}_0}{x^n}{ + ^n}{{\text{C}}_1}{x^{n - 1}}h +  \ldots { + ^n}{{\text{C}}_n}{h^n}$

${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(x + h)}^n} - {x^n}}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( {n{x^{n - 1}} +  \ldots  + {h^{n - 1}}} \right]}}{h} = n{x^{n - 1}}.$

Example 12: Find the derivative of $2{x^4} + x$.

Ans: Given:$2{x^4} + x$

Derivative of the sum of two functions is the sum of the derivatives of the functions.

$\dfrac{d}{{dx}}[f(x) + g(x)] = \dfrac{d}{{dx}}f(x) + \dfrac{d}{{dx}}g(x)$

Let $y = 2{x^4} + x$

Differentiating both sides with respect to $x$, 

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {2{x^4}} \right) + \dfrac{d}{{dx}}(x)$

$= 2 \times 4{x^{4 - 1}} + 1{x^0}$ 

$= 8{x^3} + 1 \cdot \dfrac{d}{{dx}}\left( {2{x^4} + x} \right)$

$= 8{x^3} + 1.$ 

Example 13: Find the derivative of ${x^2}cosx$.

Ans: Given:${x^2}\cos x$

Derivative of the product of two functions is given by the following product rule.

$\dfrac{d}{{dx}}[f(x) \cdot g(x)] = \dfrac{d}{{dx}}f(x) \cdot g(x) + f(x) \cdot \dfrac{d}{{dx}}g(x)$

Let $y = {x^2}\cos x$

Differentiating both sides with respect to $x$, 

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2}\cos x} \right)$

$= {x^2}\dfrac{d}{{dx}}(\cos x) + \cos x\dfrac{d}{{dx}}\left( {{x^2}} \right)$

$= {x^2}( - \sin x) + \cos x(2x)$

$= 2x\cos x - {x^2}\sin x$ 

Example 14: Evaluate $\mathop {\lim }\limits_{x \to \dfrac{\pi }{6}} \dfrac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}$.

Ans: Given: $\mathop {\lim }\limits_{x \to \dfrac{\pi }{6}} \dfrac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}$

$\mathop {\lim }\limits_{x \to \dfrac{\pi }{6}} \dfrac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}$

$= \mathop {\lim }\limits_{x \to \dfrac{\pi }{6}} \dfrac{{(2\sin x - 1)(\sin x + 1)}}{{(2\sin x - 1)(\sin x - 1)}}$

$= \mathop {\lim }\limits_{x \to \dfrac{\pi }{6}} \dfrac{{\sin x + 1}}{{\sin x - 1}}\quad ({\text{ as }}2\sin x - 1 \ne 0)$

$= \dfrac{{1 + \sin \dfrac{\pi }{6}}}{{\sin \dfrac{\pi }{6} - 1}} =  - 3$ 

Example 15: Evaluate $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x - \sin x}}{{{{\sin }^3}x}}$.

Ans: Given: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x - \sin x}}{{{{\sin }^3}x}}$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x - \sin x}}{{{{\sin }^3}x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x\dfrac{1}{{\cos x}} - 1}}{{{{\sin }^3}x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{\cos x{{\sin }^2}x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{\cos x - 4{{\sin }^2}\dfrac{x}{2} \cdot {{\cos }^2}\dfrac{x}{2}}} = \dfrac{1}{2}$ 

Example 16: Evaluate $\mathop {\lim }\limits_{x \to a} \dfrac{{\sqrt {a + 2x}  - \sqrt {3x} }}{{\sqrt {3a + x}  - 2\sqrt x }}$.

Ans: Given: $\mathop {\lim }\limits_{x \to a} \dfrac{{\sqrt {a + 2x}  - \sqrt {3x} }}{{\sqrt {3a + x}  - 2\sqrt x }}$

Let $f$ and $g$ be two functions such that both $\mathop {\lim }\limits_{x \to a} f(x)$ and $\mathop {\lim }\limits_{x \to a} g(x)$ exist. Then $\mathop {\lim }\limits_{x \to a} [f(x) - g(x)] = \mathop {\lim }\limits_{x \to a} f(x) - \mathop {\lim }\limits_{x \to a} g(x)$

$\mathop {\lim }\limits_{x \to a} \dfrac{{\sqrt {a + 2x}  - \sqrt {3x} }}{{\sqrt {3a + x}  - 2\sqrt x }}$

$= \mathop {\lim }\limits_{x \to a} \dfrac{{\sqrt {a + 2x}  - \sqrt {3x} }}{{\sqrt {3a + x}  - 2\sqrt x }} \times \dfrac{{\sqrt {a + 2x}  + \sqrt {3x} }}{{\sqrt {a + 2x}  + \sqrt {3x} }}$

$= \mathop {\lim }\limits_{x \to a} \dfrac{{a + 2x - 3x}}{{(\sqrt {3a + x}  - 2\sqrt x )(\sqrt {a + 2x}  + \sqrt {3x} )}}$

$= \mathop {\lim }\limits_{x \to a} \dfrac{{(a - x)(\sqrt {3a + x}  + 2\sqrt x )}}{{(\sqrt {a + 2x}  + \sqrt {3x} )(\sqrt {3a + x}  - 2\sqrt x )(\sqrt {3a + x}  + 2\sqrt x )}}$

$= \mathop {\lim }\limits_{x \to a} \dfrac{{(a - x)\sqrt {3a + x}  + 2\sqrt x }}{{(\sqrt {a + 2x}  + \sqrt {3x} )(3a + x - 4x)}}$

$= \dfrac{{4\sqrt a }}{{3 \times 2\sqrt {3a} }}$

$= \dfrac{2}{{3\sqrt 3 }} = \dfrac{{2\sqrt 3 }}{9}{\text{ }}{\text{. }}$ 

Example 17: Evaluate $\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos ax - \cos bx}}{{\cos cx - 1}}$.

Ans: Given: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos ax - \cos bx}}{{\cos cx - 1}}$

Let $f$ and $g$ be two functions such that both $\mathop {\lim }\limits_{x \to a} f(x)$ and $\mathop {\lim }\limits_{x \to a} g(x)$ exist. Then $\mathop {\lim }\limits_{x \to a} [f(x) - g(x)] = \mathop {\lim }\limits_{x \to a} f(x) - \mathop {\lim }\limits_{x \to a} g(x)$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{ - 2\sin \left( {\dfrac{{(a + b)}}{2}x} \right)\sin \left( {\dfrac{{(a - b)}}{2}x} \right)}}{{ - 2{{\sin }^2}(cx/2)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{{(a + b)x}}{2}} \right) \cdot \sin \left( {\dfrac{{(a - b)x}}{2}} \right)}}{{{x^2}}} \times \dfrac{{{x^2}}}{{{{\sin }^2}\dfrac{{cx}}{2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \dfrac{{(a + b)x}}{2}}}{{\dfrac{{(a + b)x}}{2} \cdot \left( {\dfrac{2}{{a + b}}} \right)}} \times \dfrac{{\sin \dfrac{{(a - b)x}}{2}}}{{\dfrac{{(a - b)x}}{2} \cdot \dfrac{2}{{a - b}}}} \times \dfrac{{{{\left( {\dfrac{{cx}}{2}} \right)}^2} \times \dfrac{4}{{{c^2}}}}}{{{{\sin }^2}\dfrac{{cx}}{2}}}$

$= \left[ {\left( {\dfrac{{a + b}}{2}} \right)\left( {\dfrac{{a - b}}{2}} \right)\left( {\dfrac{4}{{{c^2}}}} \right)\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{\sin \dfrac{{(a + b)x}}{2}}}{{\dfrac{{(a + b)x}}{2}}}} \right\} \times } \right.\left. {\mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{\sin \dfrac{{(a - b)x}}{2}}}{{\left( {\dfrac{{a - b}}{2}} \right)x}}} \right\}\mathop {\lim }\limits_{x \to 0} {{\left\{ {\dfrac{{\dfrac{{cx}}{2}}}{{\sin \dfrac{{cx}}{2}}}} \right\}}^2}} \right]$

$= \left( {\dfrac{{a + b}}{2} \times \dfrac{{a - b}}{2} \times \dfrac{4}{{{c^2}}}} \right)$

$= \dfrac{{{a^2} - {b^2}}}{{{c^2}}}$

Example 18: Evaluate $\mathop {\lim }\limits_{h \to 0} \dfrac{{{{(a + h)}^2}\sin (a + h) - {a^2}\sin a}}{h}$.

Ans: Given: $\mathop {\lim }\limits_{h \to 0} \dfrac{{{{(a + h)}^2}\sin (a + h) - {a^2}\sin a}}{h}$

Let $f$ and $g$ be two functions such that both $\mathop {\lim }\limits_{x \to a} f(x)$ and $\mathop {\lim }\limits_{x \to a} g(x)$ exist. Then $\mathop {\lim }\limits_{x \to a} [f(x) - g(x)] = \mathop {\lim }\limits_{x \to a} f(x) - \mathop {\lim }\limits_{x \to a} g(x)$

$\mathop {\lim }\limits_{h \to 0} \dfrac{{{{(a + h)}^2}\sin (a + h) - {a^2}\sin a}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{a^2} + {h^2} + 2ah} \right)[\sin a\cos h + \cos a\sin h] - {a^2}\sin a}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{{a^2}\sin a(\cos h - 1)}}{h} + \dfrac{{{a^2}\cos a\sin h}}{h} + (h + 2a)(\sin a\cos h + \cos a\sin h)} \right]$

$= \dfrac{{\mathop {\lim }\limits_{h \to 0} \dfrac{{{a^2}\sin a\left( { - 2{{\sin }^2}\dfrac{h}{2}} \right)}}{{\dfrac{{{h^2}}}{2}}}}}{ \cdot }\dfrac{h}{2} + \mathop {\lim }\limits_{h \to 0} \dfrac{{{a^2}\cos a\sin h}}{h} + \mathop {\lim }\limits_{h \to 0} (h + 2a)\sin (a + h)$

$= {a^2}\sin a \times 0 + {a^2}\cos a(1) + 2a\sin a$$= {a^2}\cos a + 2a\sin a.$

Example 19: Find the derivative of $f(x) = tan(ax + b)$, by first principle.

Ans: Given: $f(x) = \tan (ax + b)$

Since, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan (a(x + h) + b) - \tan (ax + b)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{\sin (ax + ah + b)}}{{\cos (ax + ah + b)}} - \dfrac{{\sin (ax + b)}}{{\cos (ax + b)}}}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (ax + ah + b)\cos (ax + b) - \sin (ax + b)\cos (ax + ah + b)}}{{h\cos (ax + b)\cos (ax + ah + b)}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{a\sin (ah)}}{{a \cdot h\cos (ax + b)\cos (ax + ah + b)}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{a}{{\cos (ax + b)\cos (ax + ah + b)}}\mathop {\lim }\limits_{ah \to 0} \dfrac{{\sin ah}}{{ah}}$

$= \dfrac{a}{{{{\cos }^2}(ax + b)}} = a{\sec ^2}(ax + b).$ 

$\left[{\text{ as }} h \to 0 \Rightarrow ah \to 0\right]$ 

Example 20: Find the derivative of $f(x) = \sqrt {sinx}$, by first principle.

Ans: Given: $f(x) = \sqrt {\sin x}$

Since, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {\sin (x + h)}  - \sqrt {\sin x} }}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{(\sqrt {\sin (x + h)}  - \sqrt {\sin x} )(\sqrt {\sin (x + h)}  + \sqrt {\sin x} )}}{{h(\sqrt {\sin (x + h)}  + \sqrt {\sin x} )}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (x + h) - \sin x}}{{h(\sqrt {\sin (x + h)}  + \sqrt {\sin x} )}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{2\cos \dfrac{{2x + h}}{2}\sin \dfrac{h}{2}}}{{2 \cdot \dfrac{h}{2}(\sqrt {\sin (x + h)}  + \sqrt {\sin x} )}}$

$= \dfrac{{\cos x}}{{2\sqrt {\sin x} }} = \dfrac{1}{2}\cot x\sqrt {\sin x}$

Example 21: Find the derivative of $\dfrac{{cosx}}{{1 + sinx}}$.

Ans: Given: $\dfrac{{\cos x}}{{1 + \sin x}}$

Derivative of quotient of two functions is given by the following quotient rule $\dfrac{d}{{dx}}\dfrac{{f(x)}}{{g(x)}} = \dfrac{{\dfrac{d}{{dx}}f(x) \cdot g(x) - f(x) \cdot \dfrac{d}{{dx}}g(x)}}{{{{(g(x))}^2}}}$

Assume, $y = \dfrac{{\cos x}}{{1 + \sin x}}$

Differentiating both sides with respects to $x$,

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\dfrac{{\cos x}}{{1 + \sin x}}$

$= \dfrac{{(1 + \sin x)\dfrac{d}{{dx}}(\cos x) - \cos x\dfrac{d}{{dx}}(1 + \sin x)}}{{{{(1 + \sin x)}^2}}}$

$= \dfrac{{(1 + \sin x)( - \sin x) - \cos x(\cos x)}}{{{{(1 + \sin x)}^2}}}$ 

$= \dfrac{{ - \sin x - {{\sin }^2}x - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}$

$= \dfrac{{ - (1 + \sin x)}}{{{{(1 + \sin x)}^2}}} = \dfrac{{ - 1}}{{1 + \sin x}}$

Example22: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{x(1 + \cos x)}}$ is equal to

(A) 0

(B) $\dfrac{1}{2}$

(C) 1

(D) $- 1$

Ans: Given: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{x(1 + \cos x)}}$

If $\dfrac{{{\text{f}}({\text{x}})}}{{{\text{g}}({\text{x}})}}$ is in the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ then:

$\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{{f^\prime }(x)}}{{{g^\prime }(x)}}$

Applying L’ hospital rule

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{x(1 + \cos x)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x}}{{1 + \cos x - x\sin x}}$

$= \dfrac{{\cos 0}}{{1 + \cos 0 - 0 \cdot \sin 0}}$

$= \dfrac{1}{{1 + 1 - 0}}$$= \dfrac{1}{2}$

Correct Option: B

Example 23: $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{1 - \sin x}}{{\cos x}}$ is equal to

(A) 0

(B) $- 1$

(C) 1

(D) Does not exist

Ans: Given: $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{1 - \sin x}}{{\cos x}}$

Let $f$ and $g$ be two functions such that both $\mathop {\lim }\limits_{x \to a} f(x)$ and $\mathop {\lim }\limits_{x \to a} g(x)$ exist. Then $\mathop{\lim }\limits_{x \to a} \left[ f(x) - g(x) \right]$ = $\mathop{\lim }\limits_{x \to a} f(x) - \mathop {\lim }\limits_{x \to a} g(x)$

$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{1 - \sin x}}{{\cos x}}$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{1 - \sin \left( {\dfrac{\pi }{2} - y} \right)}}{{\cos \left( {\dfrac{\pi }{2} - y} \right)}}\quad {\text{ taking }}\dfrac{\pi }{2} - x = y$ 

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{1 - \cos y}}{{\sin y}}$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{2{{\sin }^2}\dfrac{y}{2}}}{{2\sin \dfrac{y}{2}\cos \dfrac{y}{2}}}$

$= \mathop {\lim }\limits_{y \to 0} \tan \dfrac{y}{2} = 0$ 

Correct Option: A

Example 24: $\mathop {\lim }\limits_{x \to 0} \dfrac{{|x|}}{x}$ is equal to

(A) 1

(B) $- 1$

(C) 0

(D) Does not exists 

Ans: Given: $\mathop {\lim }\limits_{x \to 0} \dfrac{{|x|}}{x}$

If the right and left hand limits coincide, then the common value is considered as the limit of $f$ at $x = a$ and denote it by $\mathop {\lim }\limits_{x \to a} f(x)$.

${\text{LHL}} = \mathop {\lim }\limits_{h \to 0} f(0 - h)$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{|0 - h|}}{{0 - h}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} ( - 1) =  - 1$ 

${\text{RHL}} = \mathop {\lim }\limits_{x \to 0} f(0 + h)$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{|0 + h|}}{{0 + h}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{h}{h}$

$= \mathop {\lim }\limits_{h \to 0} (1) = 1$ 

$\therefore {\text{LHL}} \ne {\text{RHL}}$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{|x|}}{x}$ does not exist

Correct Option: D

Examples 25: $\mathop {\lim }\limits_{x \to 1} [x - 1]$, where [.] is greatest integer function, is equal to

(A) 1

(B) 2

(C) 0

(D) does not exists

Ans: Given: $\mathop {\lim }\limits_{x \to 1} [x - 1]$

If the right and left hand limits coincide, then the common value is considered as the limit of $f$ at $x = a$ and denote it by $\mathop {\lim }\limits_{x \to a} f(x)$.

${\text{R}}{\text{.H}}{\text{.L }} = \mathop {\lim }\limits_{x \to {1^ + }} [x - 1] = 0$

${\text{L}}{\text{.H}}{\text{.L}} = \mathop {\lim }\limits_{x \to {1^ - }} [x - 1] =  - 1$ 

$\therefore {\text{LHL}} \ne {\text{RHL}}$

$\Rightarrow \mathop {\lim }\limits_{x \to 1} [x - 1]$ does not exist

Correct Option: D

Example 26: $\mathop {\lim }\limits_{x \to 0} x\sin \dfrac{1}{x}$ is equals to

(A) 0

(B) 1

(C) $\dfrac{1}{2}$

(D) does not exist

Ans: Given: $\mathop {\lim }\limits_{x \to 0} x\sin \dfrac{1}{x}$

The squeeze (or sandwich) theorem states that if ${f(x)} \leqslant {g(x)} \leqslant {h(x)}$ for all numbers, and at some point $x = k$ we have $f(k) = h(k)$, then $g(k)$ must also be equal to them.

$\mathop {\lim }\limits_{x \to 0} x = 0{\text{ and }} - 1 \leqslant \sin \dfrac{1}{x} \leqslant 1,$ by Sandwich Theorem, 

$\mathop {\lim }\limits_{x \to 0} x \cdot  - 1 = \mathop {\lim }\limits_{x \to 0} x \cdot 1 = 0$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} x\sin \dfrac{1}{x} = 0$

Correct Option: A

Example 27: $\mathop {\lim }\limits_{n \to \infty } \dfrac{{1 + 2 + 3 +  \ldots  + n}}{{{n^2}}},n \in {\mathbf{N}}$, is equal to

(A) 0

(B) 1

(C) $\dfrac{1}{2}$

(D) $\dfrac{1}{4}$

Ans: Given: $\mathop {\lim }\limits_{n \to \infty } \dfrac{{1 + 2 + 3 +  \ldots  + n}}{{{n^2}}},n \in {\mathbf{N}}$

If the right and left hand limits coincide, then the common value is considered as the limit of $f$ at $x = a$ and denote it by$\mathop {\lim }\limits_{x \to a} f(x)$.

$\mathop {\lim }\limits_{x \to \infty } \dfrac{{1 + 2 + 3 +  \ldots  + n}}{{{n^2}}}$

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{{n(n + 1)}}{{2{n^2}}}$

$= \mathop {\lim }\limits_{x \to \infty } \dfrac{1}{2}\left( {1 + \dfrac{1}{n}} \right) = \dfrac{1}{2}$ 

Correct Option: C

Example 28: If $f(x) = x sinx$, then ${f’}\dfrac{\pi }{2}$ is equal to

(A) 0

(B) 1

(C) $- 1$

(D) $\dfrac{1}{2}$

Ans: Given: $f(x) = x\sin x$

The derivative of a function $f(x)$ is the function whose value at $x$ is ${f^\prime }(x)$

As ${f^\prime }(x) = x\cos x + \sin x$

${f^\prime }\left( {\dfrac{\pi }{2}} \right)$

$\Rightarrow \dfrac{\pi }{2}\cos \dfrac{\pi }{2} + \sin \dfrac{\pi }{2} = 1$ 

Correct Option: B

Exercise 13.1

Short Answer Type Question

1. Evaluate: $\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}}$

\[\because [{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)]\]

$= \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}}{\text{ }}\left( {{\text{given}}} \right)$

 $= \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - {{(3)}^2}}}{{x - 3}}$

$= \mathop {\lim }\limits_{x \to 3} \dfrac{{(x + 3)(x - 3)}}{{(x - 3)}}$

$   = \mathop {\lim }\limits_{x \to 3} (x + 3)$

$  = 3 + 3 = 6$ 

2. Evaluate: $\mathop {\lim }\limits_{x \to 1/2} \dfrac{{4{x^2} - 1}}{{2x - 1}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 1/2} \dfrac{{4{x^2} - 1}}{{2x - 1}}$

Then,

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{{{(2x)}^2} - {{(1)}^2}}}{{2x - 1}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{(2x + 1)(2x - 1)}}{{(2x - 1)}}$

$= \mathop {\lim }\limits_{x \to 1/2} (2x + 1)$

$= 2 \times \dfrac{1}{2} + 1$

$= 1 + 1$

$= 2$ 

3. Evaluate: $\mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h}  - \sqrt x }}{h}$

Ans: Given:

$\mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h}  - \sqrt x }}{h}$

\[\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}\]

\[\because h \to 0 \Rightarrow x + h \to x\]

\[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h}  - \sqrt x }}{h}{\text{ }}\left( {{\text{given}}} \right)\]

\[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(x + h)}^{1/2}} - {{(x)}^{1/2}}}}{{x + h - x}}\]

\[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(x + h)}^{1/2}} - {{(x)}^{1/2}}}}{{(x + h) - x}}\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

\[ = \dfrac{1}{2}{x^{\dfrac{1}{2} - 1}}\]

\[ = \dfrac{1}{2}{x^{ - 1/2}}\quad {\text{  }}\left[ {\because h \to 0 \Rightarrow x + h \to x} \right]\]

\[ = \dfrac{1}{{2\sqrt x }}\]

4. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(x + 2)}^{1/3}} - {2^{1/3}}}}{x}{\text{ }}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(x + 2)}^{1/3}} - {2^{1/3}}}}{x}{\text{ }}$

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(x + 2)}^{1/3}} - {2^{1/3}}}}{{(x + 2) - 2}}$

$= \dfrac{1}{3} \times {2^{\dfrac{1}{3} - 1}} = \dfrac{1}{3} \times {(2)^{ - 2/3}}\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$

$= \dfrac{1}{{3{{(2)}^{2/3}}}}\quad {\text{                    }}[\because x \to 0 \Rightarrow x + 2 \to 2]$

5. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^6} - 1}}{{{{(1 + x)}^2} - 1}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^6} - 1}}{{{{(1 + x)}^2} - 1}}$

Divide numerator and denominator by \[x\]

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{{(1 + x)}^6} - 1}}{x}}}{{\dfrac{{{{(1 + x)}^2} - 1}}{x}}}$\[\left[ {\because x \to 0 \Rightarrow 1 + x \to 1} \right]\]

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{{(1 + x)}^6} - 1}}{{(1 + x) - 1}}}}{{\dfrac{{{{(1 + x)}^2} - 1}}{{(1 + x) - 1}}}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^6} - {{(1)}^6}}}{{(1 + x) - 1}}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^2} - {{(1)}^2}}}{{(1 + x) - 1}}}}$$\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}} \right]$

$= \dfrac{{6{{(1)}^{6 - 1}}}}{{2{{(1)}^{2 - 1}}}}$\[\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

$= \dfrac{{6 \times 1}}{{2 \times 1}}$

$= \dfrac{6}{2}$

$= 3$

6. Evaluate: $\mathop {\lim }\limits_{x \to a} \dfrac{{{{(2 + x)}^{5/2}} - {{(a + 2)}^{5/2}}}}{{x - a}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to a} \dfrac{{{{(2 + x)}^{5/2}} - {{(a + 2)}^{5/2}}}}{{x - a}}$

Then,

$= \mathop {\lim }\limits_{x \to a} \dfrac{{{{(2 + x)}^{5/2}} - {{(a + 2)}^{5/2}}}}{{(2 + x) - (a + 2)}}$

$= \dfrac{5}{2}{\left( {a + 2} \right)^{\dfrac{5}{2} - 1}}\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$

$= \dfrac{5}{2}{\left( {a + 2} \right)^{\dfrac{3}{2}}}{\text{                }}\left[ {\because x \to a \Rightarrow x + 2 \to a + 2} \right]$

7. Evaluate: $\mathop {\lim }\limits_{x \to 1} \dfrac{{x - \sqrt x }}{{\sqrt x  - 1}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{x - \sqrt x }}{{\sqrt x  - 1}}$

Then,

Then,

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt x \left[ {{{\left( x \right)}^{\dfrac{7}{2}}} - 1} \right]}}{{\sqrt x  - 1}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{{{(x)}^{7/2}} - 1}}{{\sqrt x  - 1}} \cdot \mathop {\lim }\limits_{x \to 1} \sqrt x$

$\left[ {\because \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot g\left( x \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right]$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{{x^{7/2}} - 1}}{{x - 1}}}}{{\dfrac{{{{(x)}^{1/2}} - 1}}{{x - 1}}}} \cdot 1$

$= \dfrac{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{7/2}} - 1}}{{x - 1}}}}{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{{(x)}^{1/2}} - 1}}{{x - 1}}}}$$\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}} \right]$

$= \dfrac{{\dfrac{7}{2}{{(1)}^{\dfrac{7}{2} - 1}}}}{{\dfrac{1}{2}{{(1)}^{\dfrac{1}{2} - 1}}}}$

$= \dfrac{{\dfrac{7}{2}}}{{\dfrac{1}{2}}}$

$= 7$

8. Evaluate: $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{\sqrt {3x - 2}  - \sqrt {x + 2} }}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{\sqrt {3x - 2}  - \sqrt {x + 2} }}$

Then, 

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left. {\left( {{x^2} - 4} \right)\sqrt {3x - 2}  + \sqrt {x + 2} } \right)}}{{(\sqrt {3x - 2}  - \sqrt {x + 2} )(\sqrt {3x - 2}  - \sqrt {x + 2} )}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {{x^2} - 4} \right)(\sqrt {3x - 2}  + \sqrt {x + 2} )}}{{{{(\sqrt {3x - 2} )}^2} - {{(\sqrt {x + 2} )}^2}}}{\text{  }}\left[ {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {{x^2} - 4} \right)(\sqrt {3x - 2}  + \sqrt {x + 2} )}}{{(3x - 2) - (x + 2)}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {{x^2} - 4} \right)(\sqrt {3x - 2}  + \sqrt {x + 2)} }}{{3x - 2 - x - 2}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {{x^2} - 4} \right)(\sqrt {3x - 2}  + \sqrt {x + 2} )}}{{2x - 4}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x + 2)(x - 2)(\sqrt {3x - 2}  + \sqrt {x + 2} )}}{{2(x - 2)}}$

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x + 2)(\sqrt {3x - 2}  + \sqrt {x + 2} )}}{2}$

$= \dfrac{{(2 + 2)(\sqrt {6 - 2}  + \sqrt {2 + 2} )}}{2}$

$= \dfrac{{4(2 + 2)}}{2} = 8$

9. Evaluate: $\mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{{x^4} - 4}}{{{x^2} + 3\sqrt 2 x - 8}}$

Ans: Given:     

$\mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{{x^4} - 4}}{{{x^2} + 3\sqrt 2 x - 8}}$

Then, 

$\mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{{x^4} - 4}}{{{x^2} + 3\sqrt 2 x - 8}}$

$= \mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{{{\left( {{x^2}} \right)}^2} - {{(2)}^2}}}{{{x^2} + 3\sqrt 2 x - 8}}$

$= \mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{\left( {{x^2} - 2} \right)\left( {{x^2} + 2} \right)}}{{{x^2} + 4\sqrt 2 x - \sqrt 2 x - 8}}$

$= \mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{(x - \sqrt 2 )(x + \sqrt 2 )\left( {{x^2} + 2} \right)}}{{x(x + 4\sqrt {2)}  - \sqrt 2 (x + 4\sqrt 2 )}}$

$= \mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{(x - \sqrt 2 )(x + \sqrt 2 )\left( {{x^2} + 2} \right)}}{{(x - \sqrt 2 )(x + 4\sqrt 2 )}}$

$= \mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{(x + \sqrt 2 )\left( {{x^2} + 2} \right)}}{{(x + 4\sqrt 2 )}}$

\[ = \dfrac{{\left( {\sqrt 2  + \sqrt 2 } \right)\left[ {{{\left( {\sqrt 2 } \right)}^2} + 2} \right]}}{{\left( {\sqrt 2  + 4\sqrt 2 } \right)}}\]

\[ = \dfrac{{2\sqrt 2 \left( {2 + 2} \right)}}{{5\sqrt 2 }}\]

\[ = \dfrac{8}{5}\]

10. Evaluate: $\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^7} - 2{x^5} + 1}}{{{x^3} - 3{x^2} + 2}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^7} - 2{x^5} + 1}}{{{x^3} - 3{x^2} + 2}}$

Then,

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^7} - {x^5} - {x^5} + 1}}{{{x^3} - {x^2} - 2{x^2} + 2}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^5}\left( {{x^2} - 1} \right) - 1\left( {{x^5} - 1} \right)}}{{{x^2}(x - 1) - 2\left( {{x^2} - 1} \right)}}$

On dividing numerator and denominator by $\left( {x - 1} \right){\text{, then}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{{x^5}\left( {{x^2} - 1} \right)}}{{(x - 1)}} - \dfrac{{1\left( {{x^2} - 1} \right)}}{{(x - 1)}}}}{{\dfrac{{{x^2}(x - 1)}}{{(x - 1)}} - \dfrac{{2\left( {{x^2} - 1} \right)}}{{(x - 1)}}}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 1} {x^5}(x + 1) - \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{{x^5} - 1}}{{x - 1}}} \right)}}{{\mathop {\lim }\limits_{x \to 1} {x^2} - \mathop {\lim }\limits_{x \to 1} (x + 1)}}$

$= \dfrac{{1 \times 2 - 5 \times {{(1)}^4}}}{{1 - 2 \times 2}}$

$= \dfrac{{2 - 5}}{{1 - 4}}$

$= \dfrac{{ - 3}}{{ - 3}} = 1$

11. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + {x^3}}  - \sqrt {1 - {x^3}} }}{{{x^2}}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + {x^3}}  - \sqrt {1 - {x^3}} }}{{{x^2}}}$

Then, 

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + {x^3}}  - \sqrt {1 - {x^3}} }}{{{x^2}}}{\text{ }}\left( {{\text{given}}} \right)$

Rationalise with $\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + {x^3}}  - \sqrt {1 - {x^3}} }}{{{x^2}}} \cdot \dfrac{{\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} }}{{\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} }}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 + {x^3}} \right) - \left( {1 - {x^3}} \right)}}{{{x^2}\left( {\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} } \right)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + {x^3} - 1 + {x^3}}}{{{x^2}\left( {\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} } \right)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{x^3}}}{{{x^2}\left( {\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} } \right)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2x}}{{\left( {\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} } \right)}}$

$= 0$

12. Evaluate: \[\mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^3} + 27}}{{{x^5} + 243}}\]

Ans: Given:

\[\mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^3} + 27}}{{{x^5} + 243}}\]

Then, 

\[ = \mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^3} + 27}}{{{x^5} + 243}}{\text{ }}\left( {{\text{given}}} \right)\]

\[ = \mathop {\lim }\limits_{x \to  - 3} \dfrac{{\dfrac{{{x^3} + 27}}{{x + 3}}}}{{\dfrac{{{x^5} + 243}}{{x + 3}}}}\]

\[ = \mathop {\lim }\limits_{x \to  - 3} \dfrac{{\dfrac{{{x^3} - {{( - 3)}^3}}}{{x - ( - 3)}}}}{{\dfrac{{{x^5} - {{( - 3)}^5}}}{{x - ( - 3)}}}}\]

\[ = \dfrac{{\mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^3} - {{( - 3)}^3}}}{{x - ( - 3)}}}}{{\mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^5} - {{( - 3)}^5}}}{{x - ( - 3)}}}}{\text{ }}\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}} \right]\]

\[ = \dfrac{{3{{( - 3)}^{3 - 1}}}}{{5{{( - 3)}^{5 - 1}}}}\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

\[ = \dfrac{3}{5}\dfrac{{{{( - 3)}^2}}}{{{{( - 3)}^4}}}\]

\[ = \dfrac{3}{{5{{( - 3)}^2}}}\]

\[ = \dfrac{3}{{45}}\]

\[ = \dfrac{1}{{15}}\]

13. Evaluate: $\mathop {\lim }\limits_{x \to 1/2} \left( {\dfrac{{8x - 3}}{{2x - 1}} - \dfrac{{4{x^2} + 1}}{{4{x^2} - 1}}} \right)$.

Ans: Given:

$\mathop {\lim }\limits_{x \to 1/2} \left( {\dfrac{{8x - 3}}{{2x - 1}} - \dfrac{{4{x^2} + 1}}{{4{x^2} - 1}}} \right)$

Then,

$= \mathop {\lim }\limits_{x \to 1/2} \left( {\dfrac{{8x - 3}}{{2x - 1}} - \dfrac{{4{x^2} + 1}}{{4{x^2} - 1}}} \right)\,\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 1/2} \left[ {\dfrac{{(8x - 3)(2x + 1) - \left( {4{x^2} + 1} \right)}}{{\left( {4{x^2} - 1} \right)}}} \right]$

$= \mathop {\lim }\limits_{x \to 1/2} \left[ {\dfrac{{16{x^2} + 8x - 6x - 3 - 4{x^2} - 1}}{{4{x^2} - 1}}} \right]$

$= \mathop {\lim }\limits_{x \to 1/2} \left[ {\dfrac{{12{x^2} + 2x - 4}}{{4{x^2} - 1}}} \right]$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2\left( {6{x^2} + x - 2} \right)}}{{4{x^2} - 1}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2\left( {6{x^2} + 4x - 3x - 2} \right)}}{{4{x^2} - 1}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2[2x(3x + 2) - 1(3x + 2)]}}{{4{x^2} - 1}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2[(3x + 2)(2x - 1)]}}{{{{(2x)}^2} - {{(1)}^2}}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2(3x + 2)(2x - 1)}}{{(2x - 1)(2x + 1)}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2(3x + 2)}}{{2x - 1}}$

$= \dfrac{{2\left( {3 \times \dfrac{1}{2} + 2} \right)}}{{2 \times \dfrac{1}{2} + 1}}$

$= \dfrac{3}{2} + 2$

$= \dfrac{7}{2}$ 

14. Find ‘n’, if \[\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 80,{\text{ }}n \in N\]

Ans: Given:

\[\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 80,{\text{ }}n \in N\]

Then, 

\[ \Rightarrow {\text{ }}n{(2)^{n - 1}} = 80{\text{  }}\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

\[ \Rightarrow {\text{ }}n \times {2^{n - 1}} = 5 \times 16\]

\[ \Rightarrow {\text{ }}n \times {2^{n - 1}} = 5 \times {(2)^4}\]

\[ \Rightarrow {\text{ }}n \times {2^{n - 1}} = 5 \times {(2)^{5 - 1}}\]

\[\therefore n = 5\]

15. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{\sin 7x}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{\sin 7x}}$

We know that,

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$\because x \to 0 \Rightarrow \left( {kx \to 0} \right),{\text{ here }}k{\text{ is real number}}$

\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{\sin 7x}}{\text{ }}\left( {{\text{given}}} \right)\]

Multiply and divide 3x in numerator and denominator 

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 3x}}{{3x}} \cdot 3x}}{{\dfrac{{\sin 7x}}{{7x}} \cdot 7x}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 7x}}{{7x}}}} \cdot \dfrac{{3x}}{{7x}}$

$= \dfrac{3}{7} \cdot \dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 7x}}{{7x}}}}{\text{  }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= \dfrac{3}{7}{\text{                    }}\left[ {\because x \to 0 \Rightarrow \left( {kx \to 0} \right),{\text{ here }}k{\text{ is real number}}} \right]$

16. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{{{\sin }^2}4x}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{{{\sin }^2}4x}}$

We know that, \[\because [\sin 2\theta  = 2\sin \theta \cos \theta ]\]

\[\because \cos 0 = 1\]

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x\mid }}{{{{[\sin 2(2x)]}^2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{{{(2\sin 2x\cos 2x)}^2}}}$

\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{4{{\sin }^2}2x{{\cos }^2}2x}}{\text{ }}\left[ {\because \sin 2\theta  = 2\sin \theta \cos \theta } \right]\]

\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{4{{\cos }^2}2x}}\left[ {\because \cos 0 = 1} \right]\]

\[ = \dfrac{1}{4}\]

17. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}}$

We know that, $\cos 2x = 1 - 2{\sin ^2}x$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - 1 + 2{{\sin }^2}x}}{{{x^2}}}{\text{ }}\left[ {\because \cos 2x = 1 - 2{{\sin }^2}x} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}x}}{{{x^2}}}$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}x}}{{{x^2}}}$ 

$= 2\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin x}}{x}} \right)^2}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 2 \times 1 = 2$

18. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x - \sin 2x}}{{{x^3}}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x - \sin 2x}}{{{x^3}}}$

We know that,

$\because \sin 2x = 2\sin x\cos x$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x - \sin 2x}}{{{x^3}}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x - 2\sin x\cos x}}{{{x^3}}}{\text{ }}\left[ {\because \sin 2x = 2\sin x\cos x} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x(1 - \cos x)}}{{{x^3}}}$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{1 - \cos x}}{{{x^2}}}} \right)$

$= 2 \cdot 1\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{{x^2}}}{\text{                }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - 1 + 2{{\sin }^2}\dfrac{x}{2}}}{{{x^2}}}$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{4 \times \dfrac{{{x^2}}}{4}}}$ 

$= \dfrac{{2 \cdot 2}}{4}\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2}$

$= \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2} = 1$ 

19. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos mx}}{{1 - \cos nx}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos mx}}{{1 - \cos nx}}$

We know that,

$\because \cos mx = 1 - 2{\sin ^2}\dfrac{{mx}}{2}{\text{and }}\sin nx = 1 - 2{\sin ^2}\dfrac{{nx}}{2}$

\[\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\]

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos mx}}{{1 - \cos nx}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - 1 + 2{{\sin }^2}\dfrac{{mx}}{2}}}{{1 - 1 + 2{{\sin }^2}\dfrac{{nx}}{2}}}$

${\because \cos mx = 1 - 2{{\sin }^2}\dfrac{{mx}}{2}}$ 

${{\text{ and }}\sin nx = 1 - 2{{\sin }^2}\dfrac{{nx}}{2}}$

$\mathop { = \lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}\dfrac{{mx}}{2}}}{{{{\sin }^2}\dfrac{{nx}}{2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{{\sin }^2}\dfrac{{mx}}{2}}}{{{{\left( {\dfrac{{mx}}{2}} \right)}^2}}} \cdot {{\left( {\dfrac{{mx}}{2}} \right)}^2}}}{{\dfrac{{{{\sin }^2}\dfrac{{nx}}{2}}}{{{{\left( {\dfrac{{nx}}{2}} \right)}^2}}} \cdot {{\left( {\dfrac{{nx}}{2}} \right)}^2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\sin \dfrac{{mx}}{2}}}{{\dfrac{{mx}}{2}}}} \right)}^2}}}{{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\sin \dfrac{{nx}}{2}}}{{\dfrac{{nx}}{2}}}} \right)}^2}}} \cdot \dfrac{{{m^2}\dfrac{{{x^2}}}{4}}}{{{n^2}\dfrac{{{x^2}}}{4}}}$

\[ = \dfrac{{{m^2}}}{{{n^2}}} \cdot \dfrac{{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\sin \dfrac{{mx}}{2}}}{{\dfrac{{mx}}{2}}}} \right)}^2}}}{{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\sin \dfrac{{nx}}{2}}}{{\dfrac{{nx}}{2}}}} \right)}^2}}}{\text{  }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]\]

\[ = \dfrac{{{m^2}}}{{{n^2}}}{\text{                          }}\left[ {\because x \to 0 \Rightarrow kx \to 0} \right]\]

20. Evaluate: $\mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt {1 - \cos 6x} }}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt {1 - \cos 6x} }}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}$

We know that,

$\because \cos 2x = 1 - 2{\sin ^2}x$

$\because \sin \left( {\pi  - \theta } \right) = \sin \theta$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$\because x \to \dfrac{\pi }{3} \Rightarrow \left( {x - \dfrac{\pi }{3}} \right) \to 0$

Then,

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt {1 - \cos 6x} }}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt {1 - 1 + 2{{\sin }^2}3x} }}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}{\text{ }}\left[ {\because \cos 2x = 1 - 2{{\sin }^2}x} \right]$

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt 2 \sin 3x}}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sin 3x}}{{\dfrac{\pi }{3} - x}}$ 

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sin (\pi  - 3x)}}{{\dfrac{{\pi  - 3x}}{3}}}{\text{         }}\left[ {\because \sin \left( {\pi  - \theta } \right) = \sin \theta } \right]$

$= 3\mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sin (\pi  - 3x)}}{{(\pi  - 3x)}}{\text{       }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 3 \times 1$ 

$= 3{\text{  }}\left[ {\because x \to \dfrac{\pi }{3} \Rightarrow \left( {x - \dfrac{\pi }{3}} \right) \to 0} \right]$

21. Evaluate: $\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\left( {\sin x - \cos x} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}{\text{   }}$

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\left( {\sin x - \cos x} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}{\text{   }}$

We know that,

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$\because x \to \dfrac{\pi }{4} \Rightarrow \left( {x - \dfrac{\pi }{4}} \right) \to 0$

Then,

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\left( {\sin x - \cos x} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sqrt 2 \left( {\sin x \cdot \dfrac{1}{{\sqrt 2 }} - \cos x \cdot \dfrac{1}{{\sqrt 2 }}} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sqrt 2 \left( {\sin x\cos \dfrac{\pi }{4} - \cos x \cdot \sin \dfrac{\pi }{4}} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sqrt 2 \left\{ {\sin \left( {x - \dfrac{\pi }{4}} \right)} \right\}}}{{\left( {x - \dfrac{\pi }{4}} \right)}}$

$= \sqrt 2 \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}{\text{  }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= \sqrt 2 \quad {\text{                          }}\left[ {\because x \to \dfrac{\pi }{4} \Rightarrow \left( {x - \dfrac{\pi }{4}} \right) \to 0} \right]$

22. Evaluate: $\mathop {\lim }\limits_{x \to \pi /6} \dfrac{{\sqrt 3 \sin x - \cos x}}{{x - \dfrac{\pi }{6}}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /6} \dfrac{{\sqrt 3 \sin x - \cos x}}{{x - \dfrac{\pi }{6}}}$

We know that,

$\because \sin A\cos B - \cos A\sin B = \sin (A - B)]$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}x \to \dfrac{\pi }{6} \Rightarrow \left( {x - \dfrac{\pi }{6}} \right) \to 0$

Then,

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{\sqrt 3 \sin x - \cos x}}{{x - \dfrac{\pi }{6}}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{2\left( {\dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x} \right)}}{{\left( {x - \dfrac{\pi }{6}} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{2\left( {\sin x\cos \dfrac{\pi }{6} - \cos x\sin \dfrac{\pi }{6}} \right)}}{{\left( {x - \dfrac{\pi }{6}} \right)}}{\text{  }}[\because \sin A\cos B - \cos A\sin B = \sin (A - B)]$

$= 2\mathop {\lim }\limits_{x \to \pi /6} \dfrac{{\sin \left( {x - \dfrac{\pi }{6}} \right)}}{{\left( {x - \dfrac{\pi }{6}} \right)}}$

$\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$\left[ {\because x \to \dfrac{\pi }{6} \Rightarrow \left( {x - \dfrac{\pi }{6}} \right) \to 0} \right]$ 

$= 2$

23. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x + 3x}}{{2x + \tan 3x}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x + 3x}}{{2x + \tan 3x}}$

We know that,

\[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\,{\text{and }}\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x + 3x}}{{2x + \tan 3x}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 2x + 3x}}{{2x}} \cdot 2x}}{{\dfrac{{2x + \tan 3x}}{{3x}} \cdot 3x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{\sin 2x}}{{2x}} + \dfrac{{3x}}{{2x}}} \right)2x}}{{\left( {\dfrac{{2x}}{{3x}} + \dfrac{{\tan 3x}}{{3x}}} \right)3x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 2x}}{{2x}} + \dfrac{3}{2}}}{{\dfrac{2}{3} + \dfrac{{\tan 3x}}{{3x}}}} \cdot \dfrac{2}{3}$

$= \dfrac{2}{3}\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 2x}}{{2x}} + \dfrac{3}{2}}}{{\dfrac{2}{3} + \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 3x}}{{3x}}}}$

\[ = \dfrac{2}{3}\left( {\dfrac{{1 + \dfrac{3}{2}}}{{\dfrac{2}{3} + 1}}} \right)\quad \left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right.\,{\text{and }}\left. {\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1} \right]\]

$= \dfrac{2}{3} \times \dfrac{{\dfrac{5}{5}}}{{\dfrac{5}{3}}} = \dfrac{2}{3} \times \dfrac{5}{2} \times \dfrac{3}{5}$

$= 1$ 

24. Evaluate: $\mathop {\lim }\limits_{x \to a} \dfrac{{\sin x - \sin a}}{{\sqrt x  - \sqrt a }}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to a} \dfrac{{\sin x - \sin a}}{{\sqrt x  - \sqrt a }}$

We know that,

$\because \sin C - \sin D = 2\cos \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

Then,

\[ = \mathop {\lim }\limits_{x \to a} \dfrac{{\sin x - \sin a}}{{\sqrt x  - \sqrt a }}{\text{ }}\left( {{\text{given}}} \right)\]

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{\sqrt x  - \sqrt a }}{\text{ }}\left[ {\because \sin C - \sin D = 2\cos \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}} \right]$

$= \mathop {\lim }\limits_{x \to a} \dfrac{{2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)(\sqrt {x + \sqrt a } )}}{{(\sqrt {x - \sqrt a )} (\sqrt x  + \sqrt a )}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)(\sqrt x  + \sqrt a )}}{{x - a}}$

$= 2\mathop {\lim }\limits_{x \to a} \cos \left( {\dfrac{{x + a}}{2}} \right)(\sqrt x  + \sqrt a )\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{2\left( {\dfrac{{x - a}}{2}} \right)}}$

$= 2\mathop {\lim }\limits_{x \to 0} \cos \left( {\dfrac{{x + a}}{2}} \right)(\sqrt x  + \sqrt a ) \cdot \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{\left( {\dfrac{{x - a}}{2}} \right)}}$

$= 2 \cdot \cos \dfrac{a}{2} \cdot \sqrt a  \cdot \dfrac{1}{2}\quad {\text{                 }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= \sqrt a \cos \dfrac{a}{2}$

25. Evaluate:  $\mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\cot }^2}x - 3}}{{\operatorname{cosec} x - 2}}$.

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\cot }^2}x - 3}}{{\operatorname{cosec} x - 2}}$

We know that,

$\because {\operatorname{cosec} ^2}x = 1 + {\cot ^2}x$

Then,

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\cot }^2}x - 3}}{{\operatorname{cosec} x - 2}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\operatorname{cosec} }^2}x - 1 - 3}}{{\operatorname{cosec} x - 2}}\quad \left[ {\because {{\operatorname{cosec} }^2}x = 1 + {{\cot }^2}x} \right]$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\operatorname{cosec} }^2}x - 4}}{{\operatorname{cosec} x - 2}}$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{(\operatorname{cosec} x)}^2} - {{(2)}^2}}}{{\operatorname{cosec} x - 2}}$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{(\operatorname{cosec} x + 2)(\operatorname{cosec} x - 2)}}{{(\operatorname{cosec} x - 2)}}$

$= \mathop {\lim }\limits_{x \to \pi /6} (\operatorname{cosec} x + 2)$

$= \operatorname{cosec} \dfrac{\pi }{6} + 2 = 2 + 2$

$= 4$

26. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2  - \sqrt {1 + \cos x} }}{{{{\sin }^2}x}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2  - \sqrt {1 + \cos x} }}{{{{\sin }^2}x}}$

We know that,

$\because \cos x = 2{\cos ^2}\dfrac{x}{2} - 1$

$\because \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2  - \sqrt {1 + 2{{\cos }^2}\dfrac{x}{2} - 1} }}{{{{\sin }^2}x}}\quad \left[ {\because \cos x = 2{{\cos }^2}\dfrac{x}{2} - 1} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2  - \sqrt {2{{\cos }^2}\dfrac{x}{2}} }}{{{{\sin }^2}x}}\quad {\text{        }}\left[ {\because \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2 \left( {1 - \cos \dfrac{x}{2}} \right)}}{{{{\sin }^2}x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2 \left( {1 - 1 + 2{{\sin }^2}\dfrac{x}{4}} \right)}}{{{{\sin }^2}x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2 \left( {2{{\sin }^2}\dfrac{x}{4}} \right)}}{{{{\sin }^2}x}}$

$= \mathop {\lim }\limits_{x \to 0} 2\sqrt 2 \dfrac{{{{\sin }^2}\dfrac{x}{4}}}{{{{\left( {\dfrac{x}{4}} \right)}^2}}} \cdot \dfrac{{{{\left( {\dfrac{x}{4}} \right)}^2}}}{{{{\sin }^2}x}}$

$= 2\sqrt 2 \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{4}}}{{\dfrac{x}{4}}}} \right)^2} \cdot \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{x}{{\sin x}}} \right)^2} \cdot \dfrac{1}{{16}}$

$= 2\sqrt 2  \cdot 1 \cdot 1 \cdot \dfrac{1}{{16}}$

$= \dfrac{1}{{4\sqrt 2 }}$

27. Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - 2\sin 3x + \sin 5x}}{x}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - 2\sin 3x + \sin 5x}}{x}$

We know that,

$\because \sin x + \sin y = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

Then,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - 2\sin 3x + \sin 5x}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 5x + \sin x - 2\sin 3x}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin 3x\cos 2x - 2\sin 3x}}{x}$$\left[ {\because \sin x + \sin y = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin 3x(\cos 2x - 1)}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin 3x}}{{\dfrac{1}{3} \times 3x}}(\cos 2x - 1)$

$= 6\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}}(\cos 2x - 1)$

$= 6\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}} \cdot \mathop {\lim }\limits_{x \to 0} (\cos 2x - 1)$$\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 6 \times 1 \times 0 = 0$

28. Evaluate: If $\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$, then find the value of \[k\].

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$

We know that,

$\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$

$\because \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}$

Then,

$\Rightarrow \quad \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}\,\left( {{\text{given}}} \right)$

$\Rightarrow \quad 4{(1)^{4 - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{\dfrac{{{x^3} - {k^3}}}{{x - k}}}}{{\dfrac{{{x^2} - {k^2}}}{{x - k}}}}\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$

$\Rightarrow \quad 4 = \dfrac{{\mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{x - k}}}}{{\mathop {\lim }\limits_{x \to k} \dfrac{{{x^2} - {k^2}}}{{x - k}}}}1\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}} \right]$

$\Rightarrow \quad 4 = \dfrac{{3{k^2}}}{{2k}}$

$\Rightarrow {\text{ }}4 = \dfrac{3}{2}k$

$\therefore \quad k = \dfrac{{4 \times 2}}{3} = \dfrac{8}{3}$

Differentiate each of the functions w. r. t x  in Exercises 29 to 42:

29. \[\dfrac{{{x^4} + {x^3} + {x^2} + 1}}{x}\]

Ans: Given:

\[\dfrac{{{x^4} + {x^3} + {x^2} + 1}}{x}\]

Then

\[ = \dfrac{d}{{dx}}\left( {\dfrac{{{x^4} + {x^3} + {x^2} + 1}}{x}} \right)\]

\[ = \dfrac{d}{{dx}}\left( {{x^3} + {x^2} + x + \dfrac{1}{x}} \right)\]

\[ = \dfrac{d}{{dx}}{x^3} + \dfrac{d}{{dx}}{x^2} + \dfrac{d}{{dx}}x + \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)\]          (by chain rule)

\[ = 3{x^2} + 2x + 1 + \left( { - \dfrac{1}{{{x^2}}}} \right)\]

\[ = 3{x^2} + 2x + 1 - \dfrac{1}{{{x^2}}}\]

\[ = \dfrac{{3{x^4} + 2{x^3} + {x^2} - 1}}{{{x^2}}}\]

30: ${\left( {x + \dfrac{1}{x}} \right)^3}$

Ans: Given:

${\left( {x + \dfrac{1}{x}} \right)^3}$

${\text{Let }}\quad y = {\left( {x + \dfrac{1}{x}} \right)^3}$

Differentiate on both side w.r.t \[x\]

$\therefore \quad \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {x + \dfrac{1}{x}} \right)^3}$

$= 3{\left( {x + \dfrac{1}{x}} \right)^{3 - 1}}\dfrac{d}{{dx}}\left( {x + \dfrac{1}{x}} \right)$ (by chain rule)

$= 3{\left( {x + \dfrac{1}{x}} \right)^2}\left( {1 - \dfrac{1}{{{x^2}}}} \right)$

$= 3{x^2} - \dfrac{3}{{{x^2}}} - \dfrac{3}{{{x^4}}} + 3$

31. $(3x + 5)(1 + tanx)$

Ans: Given:

$(3x + 5)(1 + \tan x)$

We know that,

Differentiate the given equation w.r.t \[x\] using product rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

Then,$\quad y = (3x + 5)(1 + \tan x)$

Differentiate on both side w.r.t \[x\]

$\therefore \quad \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}[(3x + 5)(1 + \tan x)]$

$= (3x + 5)\dfrac{d}{{dx}}(1 + \tan x) + (1 + \tan x)\dfrac{d}{{dx}}(3x + 5){\text{ }}\left[ {{\text{product rule}}} \right]$

$= (3x + 5)\left( {{{\sec }^2}x} \right) + (1 + \tan x) \cdot 3$

$= (3x + 5){\sec ^2}x + 3(1 + \tan x)$

$= 3x{\sec ^2}x + 5{\sec ^2}x + 3\tan x + 3$

32. $(secx - 1)(secx + 1)$

Ans: Given:

$(\sec x - 1)(\sec x + 1)$

We know that, Differentiate the given equation w.r.t \[x\] using chain rule

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

$\because (a + b)(a - b) = {a^2} - {b^2}$

Let, $y = (\sec x - 1)(\sec x + 1)$

$y = \left( {{{\sec }^2} - 1} \right)$$\left[ {\because (a + b)(a - b) = {a^2} - {b^2}} \right]$

$= {\tan ^2}x$

$\therefore \quad \dfrac{{dy}}{{dx}} = 2\tan x \cdot \dfrac{d}{{dx}}\tan x$

$= 2\tan x \cdot {\sec ^2}x$                    (By chain rule)

33. $\dfrac{{3x + 4}}{{5{x^2} - 7x + 9}}$

Ans: Given:

$\dfrac{{3x + 4}}{{5{x^2} - 7x + 9}}$

We know that,

Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}\]

Let,  $\quad y = \dfrac{{3x + 4}}{{5{x^2} - 7x + 9}}$

$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{\left( {5{x^2} - 7x + 9} \right)\dfrac{d}{{dx}}(3x + 4) - (3x + 4)\dfrac{d}{{dx}}\left( {5{x^2} - 7x + 9} \right)}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}{\text{ }}\left[ {{\text{by quotient rule}}} \right]$

${\text{       }} = \dfrac{{\left( {5{x^2} - 7x + 9} \right) \cdot 3 - (3x + 4)(10x - 7)}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}$

${\text{       }} = \dfrac{{15{x^2} - 21x + 27 - 30{x^2} + 21x - 40x + 28}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}$

${\text{       }} = \dfrac{{ - 15{x^2} - 40x + 55}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}$

${\text{       }} = \dfrac{{55 - 15{x^2} - 40x}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}$

34. $\dfrac{{{x^5} - cosx}}{{sinx}}$

Ans: Given:

$\dfrac{{{x^5} - \cos x}}{{\sin x}}$

We know that, Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}\]

Then,Let $y = \dfrac{{{x^5} - \cos x}}{{\sin x}}$

$\dfrac{{dy}}{{dx}} = \dfrac{{\sin x\dfrac{d}{{dx}}\left( {{x^5} - \cos x} \right) - \left( {{x^5} - \cos x} \right)\dfrac{{dy}}{{dx}}\sin x}}{{{{\left( {\sin x} \right)}^2}}}$    (by quotient rule)

$= \dfrac{{\sin x\left( {5{x^4} + \sin x} \right) - \left( {{x^5} - \cos x} \right)\cos x}}{{{{\sin }^2}x}}$

$= \dfrac{{5{x^4}\sin x + {{\sin }^2}x - {x^5}\cos x + {{\cos }^2}x}}{{{{\sin }^2}x}}$

$= \dfrac{{5{x^4}\sin x - {x^5}\cos x + {{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}}$

$= \dfrac{{5{x^4}\sin x - {x^5}\cos x + 1}}{{{{\sin }^2}x}}$

35. \[\dfrac{{{x^2}cos\dfrac{\pi }{4}}}{{sinx}}\]

Ans: Given:

\[\dfrac{{{x^2}\cos \dfrac{\pi }{4}}}{{\sin x}}\]

We know that, Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}\]

Let \[y = \dfrac{{{x^2}\cos \dfrac{\pi }{4}}}{{\sin x}}\]

\[y = \dfrac{1}{{\sqrt 2 }} \cdot \dfrac{{{x^2}}}{{\sin x}}\]

$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt 2 }}\left[ {\dfrac{{\sin x \cdot \dfrac{d}{{dx}}{x^2} - {x^2}\dfrac{d}{{dx}}\sin x}}{{{{\sin }^2}x}}} \right]$       (by quotient rule)

$= \dfrac{1}{{\sqrt 2 }}\left[ {\dfrac{{\sin x \cdot 2x - {x^2} \cdot \cos x}}{{{{\sin }^2}x}}} \right]$

$= \dfrac{1}{{\sqrt 2 }} \cdot \dfrac{{2x\sin x - {x^2}\cos x}}{{{{\sin }^2}x}}$

$= \dfrac{x}{{\sqrt 2 }}[2\operatorname{cosec} x - x\cot x\operatorname{cosec} x]$

$= \dfrac{x}{{\sqrt 2 }}\operatorname{cosec} [2 - x\cot x]$

36. \[\left( {a{x^2} + cotx} \right)\left( {p + qcosx} \right)\]

Ans: Given:

\[\left( {a{x^2} + \cot x} \right)\left( {p + q\cos x} \right)\]

We know that, Differentiate the given equation w.r.t \[x\] using product rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u\]

Let \[y = \left( {a{x^2} + \cot x} \right)\left( {p + q\cos x} \right)\]

$\therefore \quad \dfrac{{dy}}{{dx}} = \left( {a{x^2} + \cot x} \right)\dfrac{d}{{dx}}(p + q\cos x) + (p + q\cos x)\dfrac{d}{{dx}}\left( {a{x^2} + \cot x} \right){\text{ }}\left[ {{\text{by product rule}}} \right]$

$= \left( {a{x^2} + \cot x} \right)( - q\sin x) + (p + q\cos x)\left( {2ax - {{\operatorname{cosec} }^2}x} \right)$

$=  - q\sin x\left( {a{x^2} + \cot x} \right) + (p + q\cos x)\left( {2ax - {{\operatorname{cosec} }^2}x} \right)$

37. $\dfrac{{a + bsinx}}{{c + dcosx}}$

Ans: Given:

$\dfrac{{a + b\sin x}}{{c + d\cos x}}$

We know that, Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}\]

Let, $y = \dfrac{{a + b\sin x}}{{c + d\cos x}}$

$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{(c + d\cos x)\dfrac{d}{{dx}}(a + b\sin x) - (a + b\sin x)\dfrac{d}{{dx}}(c + d\cos x)}}{{{{(c + d\cos x)}^2}}}{\text{ }}\left[ {{\text{by quotient rule}}} \right]$

$= \dfrac{{(c + d\cos x)(b\cos x) - (a + b\sin x)( - d\sin x)}}{{{{(c + d\cos x)}^2}}}$

$= \dfrac{{bc\cos x + bd{{\cos }^2}x + ad\sin x + bd{{\sin }^2}x}}{{{{(c + d\cos x)}^2}}}$

$= \dfrac{{bc\cos x + ad\sin x + bd\left( {{{\cos }^2}x + {{\sin }^2}x} \right)}}{{{{(c + d\cos x)}^2}}}$

$= \dfrac{{bc\cos x + ad\sin x + bd}}{{{{(c + d\cos x)}^2}}}$

38. ${\left( {sinx + cosx} \right)^2}$

Ans: Given:

${\left( {\sin x + \cos x} \right)^2}$

We know that, Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

\[\because \cos 2x = {\cos ^2}x - {\sin ^2}x\]

\[{\text{Let }}y = {\left( {\sin x + \cos x} \right)^2}\]

\[\therefore \dfrac{{dy}}{{dx}} = 2\left( {\sin x + \cos x} \right)\left( {\sin x - \cos x} \right)\](By chain rule)

\[{\text{       }} = 2\left( {{{\cos }^2}x - {{\sin }^2}x} \right){\text{                   }}\left[ {\because \cos 2x = {{\cos }^2}x - {{\sin }^2}x} \right]\]

\[{\text{       }} = 2\cos 2x\]

39. ${(2x - 7)^2}{(3x + 5)^3}$

Ans: Given:

${(2x - 7)^2}{(3x + 5)^3}$

We know that, Differentiate the given equation w.r.t \[x\] using product rule and chain rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\] and \[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

Let $y = {(2x - 7)^2}{(3x + 5)^3}$

$\therefore \dfrac{{dy}}{{dx}} = {(2x - 7)^2}\dfrac{d}{{dx}}{(3x + 5)^3} + {(3x + 5)^3}\dfrac{d}{{dx}}{(2x - 7)^2}{\text{   }}\left[ {{\text{by product rule}}} \right]$

$= {(2x - 7)^2}(3){(3x + 5)^2}(3) + {(3x + 5)^3}2(2x - 7)(2){\text{   }}\left[ {{\text{by chain rule}}} \right]$

$= 9{(2x - 7)^2}{(3x + 5)^2} + 4{(3x + 5)^3}(2x - 7)$

$= (2x - 7){(3x + 5)^2}[9(2x - 7) + 4(3x + 5)]$

$= (2x - 7){(3x + 5)^2}(18x - 63 + 12x + 20)$

$= (2x - 7){(3x + 5)^2}(30x - 43)$

40. ${x^2}sinx + cos2x$

Ans: Given:

${x^2}\sin x + \cos 2x$

We know that, Differentiate the given equation w.r.t \[x\] using product rule 

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

${\text{Let }}y = {x^2}\sin x + \cos 2x$

$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{x^2}\sin x} \right)}}{{dx}} + \dfrac{{d\left( {\cos 2x} \right)}}{{dx}}$

${\text{       }} = \left( {2x\sin x + {x^2} \cdot \cos x} \right) + \left( { - \sin 2x \cdot 2} \right){\text{ }}\left[ {{\text{By product rule}}} \right]$

${\text{       }} = 2x\sin x + {x^2} \cdot \cos x - 2\sin 2x\left[ {{\text{By chain rule}}} \right]{\text{       }}$ 

41. \[si{n^3}xco{s^3}x\]

Ans: Given:

\[{\sin ^3}x{\cos ^3}x\]

We know that, Differentiate the given equation w.r.t \[x\] using product rule and chain rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

$\because \sin 2x = 2\sin x\cos x$

Let \[y = {\sin ^3}x{\cos ^3}x\]

\[\dfrac{{dy}}{{dx}} = {\sin ^3}x \cdot \dfrac{d}{{dx}}{\cos ^3}x + {\cos ^3}x\dfrac{d}{{dx}}{\sin ^3}x\]                    (by product rule)

$= {\sin ^3}x \cdot 3{\cos ^2}x( - \sin x) + {\cos ^3}x \cdot 3{\sin ^2}x\cos x\quad$ (by chain rule)

$=  - 3{\cos ^2}x{\sin ^4}x + 3{\sin ^2}x{\cos ^4}x$

$= 3{\sin ^2}x{\cos ^2}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)$

$= 3{\sin ^2}x{\cos ^2}x\cos 2x$

$= \dfrac{3}{4}{(2\sin x\cos x)^2}\cos 2x{\text{                                       }}\left[ {\because \sin 2x = 2\sin x\cos x} \right]$

$= \dfrac{3}{4}{\sin ^2}2x\cos 2x$

42. $\dfrac{1}{{a{x^2} + bx + c}}$

Ans: Given:

$\dfrac{1}{{a{x^2} + bx + c}}$

We know that, Differentiate the given equation w.r.t \[x\] using product rule and chain rule

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

Let $y = \dfrac{1}{{a{x^2} + bx + c}} = {\left( {a{x^2} + bx + c} \right)^{ - 1}}$

$\dfrac{{dy}}{{dx}} =  - {\left( {a{x^2} + bx + c} \right)^{ - 2}}\left( {2ax + b} \right){\text{ }}\left[ {{\text{By chain rule}}} \right]$

$= \dfrac{{ - (2ax + b)}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}}$

Long Answer Type:

Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using the first principle method.

43. \[cos\left( {{x^2} + 1} \right)\]

Ans: Given:

\[\cos \left( {{x^2} + 1} \right)\]

We know that, To differentiate the given equation

$\because \cos C - \cos D =  - 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}$

$\because x \to 0 \Rightarrow kx \to 0$

Let,

\[\therefore \dfrac{d}{{dx}}f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left\{ {{{(x + h)}^2} + 1} \right\} - \cos \left( {{x^2} + 1} \right)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2\sin \left\{ {\dfrac{{{{(x + h)}^2} + 1 + {x^2} + 1}}{2}} \right\}\sin \left\{ {\dfrac{{{{(x + h)}^2} + 1 - {x^2} - 1}}{2}} \right\}}}{h}\left[ {\because \cos C - \cos D =  - 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}} \right]$$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ { - 2\sin \left\{ {\dfrac{{{{(x + h)}^2} + {x^2} + 2}}{2}} \right\}\sin \left\{ {\dfrac{{{{(x + h)}^2} - {x^2}}}{2}} \right\}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ { - 2\sin \left\{ {\dfrac{{{{(x + h)}^2} + {x^2} + 2}}{2}} \right\}\sin \left\{ {\dfrac{{{x^2} + {h^2} + 2xh - {x^2}}}{2}} \right\}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ { - 2\sin \left\{ {\dfrac{{{{(x + h)}^2} + {x^2} + 2}}{2}} \right\}\sin \left\{ {\dfrac{{{h^2} + 2hx}}{2}} \right\}} \right]$

$=  - 2\mathop {\lim }\limits_{h \to 0} \sin \left\{ {\dfrac{{{{(x + h)}^2} + {x^2} + 2}}{2}} \right\}\mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{\sin h\left( {\dfrac{{h + 2x}}{2}} \right)}}{{h\left( {\dfrac{{h + 2x}}{2}} \right)}}} \right\} \times \left( {\dfrac{{h + 2x}}{2}} \right)$

$=  - 2\mathop {\lim }\limits_{h \to 0} \sin \left\{ {\dfrac{{{{(x + h)}^2} + {x^2} + 2}}{2}} \right\}\mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{h + 2x}}{2}} \right)\quad \left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

\[ =  - 2x\sin \left( {{x^2} + 1} \right)\quad \left[ {\because x \to 0 \Rightarrow kx \to 0} \right]\]

44. $\dfrac{{ax + b}}{{cx + d}}$

Ans: Given, $\dfrac{{ax + b}}{{cx + d}}$

Let,

 $f\left( x \right) = \dfrac{{ax + b}}{{cx + d}}$

$f\left( {x + h} \right) = \dfrac{{a\left( {x + h} \right) + b}}{{c\left( {x + h} \right) + d}}$

$\therefore \dfrac{d}{{dx}}f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {f\left( {x + h} \right) - f\left( x \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{a(x + h) + b}}{{c(x + h) + d}} - \dfrac{{ax + b}}{{cx + d}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{ax + b + ah}}{{c(x + h) + d}} - \dfrac{{ax + b}}{{cx + d}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{(ax + ah + b)(cx + d) - (ax + b\{ c(x + h) + d\} }}{{\{ c(x + h) + d\} (cx + d)}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{(ax + ah + b)(cx + d) - (ax + b)(cx + ch + d)}}{{\{ c(x + h) + d)\} (cx + d)}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{ac{x^2} + achx + bcx + adx + adh + bd - ac{x^2} - achx - adx - bcx - bch - bd}}{{\{ c(x + h) + d\} (cx + d)}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{adh - bch}}{{\{ c(x + h) + d\} (cx + d)}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{ad - bc}}{{\{ c(x + h) + d\} (cx + d)}}$

$= \dfrac{{ad - bc}}{{{{(cx + d)}^2}}}$

45. ${x^{\dfrac{2}{3}}}$

Ans: Given:

${x^{\dfrac{2}{3}}}$

We know that,

\[\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^x} - {a^n}}}{{x - a}} = n{a^{n - 1}}\]

Let,

\[f(x) = {x^{2/3}}\]

\[f(x + h) = {(x + h)^{2/3}}\]

\[\dfrac{d}{{dx}}f(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]

\[ = \mathop {\lim }\limits_{(h \to 0} \left[ {\dfrac{{(x + h)2/3 - {x^{2/3}}}}{h}} \right]\]

\[ = \mathop {\lim }\limits_{(x + h) \to x} \left[ {\dfrac{{(x + h)2/3 - {x^{2/3}}}}{{(x + h) - x}}} \right]\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^x} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

\[ = \dfrac{2}{3}{(x)^{2/3 - 1}}\quad \]

\[ = \dfrac{2}{3}{x^{ - 1/3}}\]

46. \[xcosx\]

Ans: Given:

\[x\cos x\]

We know that,

$\because \cos C - \cos D =  - 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}$

Then,

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}[(x + h)\cos (x + h) - x\cos x]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}[x\cos (x + h) + h\cos (x + h) - x\cos x]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}[x\{ \cos (x + h) - \cos x\}  + h\cos (x + h)]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {x\left\{ { - 2\sin \left( {\dfrac{{2x + h}}{2}} \right)\sin \dfrac{h}{2}} \right\} + h\cos (x + h)} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ { - 2x\sin \left( {x + \dfrac{h}{2}} \right)\dfrac{{\sin \dfrac{h}{2}}}{h} + \cos (x + h)} \right]\left[ {\because \cos C - \cos D =  - 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}} \right]$

\[ =  - 2\mathop {\lim }\limits_{h \to 0} x\sin \left( {x + \dfrac{h}{2}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \dfrac{h}{2}}}{{\dfrac{h}{2}}} \cdot \dfrac{1}{2} + \mathop {\lim }\limits_{h \to 0} \cos (x + h)\]

\[ =  - 2 \cdot \dfrac{1}{2}x\sin x + \cos x\]

\[ = \cos x - x\sin x\]

Evaluate each of the following limits in Exercises 47 to 53:

47. $\mathop{\lim }\limits_{y \rightarrow 0} \dfrac{(x+y) \sec (x+y)-x \sec x}{y}$

Ans: Given:

$\mathop {\lim }\limits_{y \to 0} \dfrac{{\left( {x + y} \right){\text{sec}}\left( {x + y} \right) - x\sec x}}{y}$

Let,

$\mathop {\lim }\limits_{y \to 0} \dfrac{{\left( {x + y} \right){\text{sec}}\left( {x + y} \right) - x\sec x}}{y}$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{\dfrac{{x + y}}{{\cos (x + y)}} - \dfrac{x}{{\cos x}}}}{y}$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{(x + y)\cos x - x\cos (x + y)}}{{y\cos x\cos (x + y)}}$

$= \mathop {\lim }\limits_{y \to 0} \left[ {\dfrac{{x\cos x + y\cos x - x\cos (x + y)}}{{y\cos x\cos (x + y)}}} \right]$

$= \mathop {\lim }\limits_{y \to 0} \left[ {\dfrac{{x\cos x - x\cos (x + y) + y\cos x}}{{y\cos x\cos (x + y)}}} \right]$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{x\{ \cos x - \cos (x + y)\}  + y\cos x}}{{y\cos x\cos (x + y)}}$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{x\left[ { - 2\sin \left( {x + \dfrac{y}{2}} \right)\sin \left( {\dfrac{{ - y}}{2}} \right)} \right] + y\cos x}}{{y\cos x\cos (x + y)}}\left[ {\because \cos C - \cos D = 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}} \right]$

$= \mathop {\lim }\limits_{y \to 0} \left[ {\dfrac{{x\left\{ {2\sin \left( {x + \dfrac{y}{2}} \right)\sin \dfrac{y}{2}} \right\} + y\cos x}}{{y\cos x\cos (x + y)}}} \right]$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{2x\sin \left( {x + \dfrac{y}{2}} \right)}}{{\cos x\cos (x + y)}} \cdot \mathop {\lim }\limits_{y \to 0} \dfrac{{\sin \dfrac{y}{2}}}{{\dfrac{y}{2}}} \cdot \dfrac{1}{2} + \mathop {\lim }\limits_{y \to 0} \sec (x + y)$

$\quad \left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right.$ and $\left. {x \to 0 \Rightarrow kx \to 0} \right]$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{2x\sin \left( {x + \dfrac{y}{2}} \right)}}{{\cos x\cos (x + y)}} \cdot \dfrac{1}{2} + \mathop {\lim }\limits_{y \to 0} \sec (x + y)$

$= \dfrac{{2x\sin x}}{{\cos x\cos x}} \cdot \dfrac{1}{2} + \sec x$

$= x\tan x\sec x + \sec x$

$= \sec x(x\tan x + 1)$

48. $\mathop{\lim }\limits_{x \rightarrow 0} \dfrac{(\sin (\alpha+\beta) x+\sin (\alpha-\beta) x+\sin 2 \alpha x)}{\cos 2 \beta x-\cos 2 \alpha x} \cdot x$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {\sin \left( {\alpha  + \beta } \right)x + \sin \left( {\alpha  - \beta } \right)x + \sin 2\alpha x} \right]}}{{\cos 2\beta x - \cos 2\alpha x}} \cdot x$

We know that,

$\because \sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$

$\because \cos C - \cos D = 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{D - C}}{2}$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}x \to 0 \Rightarrow kx \to 0{\text{ }}$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {\sin \left( {\alpha  + \beta } \right)x + \sin \left( {\alpha  - \beta } \right)x + \sin 2\alpha x} \right]}}{{\cos 2\beta x - \cos 2\alpha x}} \cdot x$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {\sin \left( {\alpha  + \beta } \right)x + \sin \left( {\alpha  - \beta } \right)x + \sin 2\alpha x} \right]}}{{\cos 2\beta x - \cos 2\alpha x}} \cdot x$

$\left[ {\because \sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {2\sin \alpha x \cdot \cos \beta x + \sin 2\alpha x} \right]}}{{\cos 2\beta x - \cos 2\alpha x}} \cdot x$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{[2\sin \alpha x\cos \beta x + \sin 2\alpha x]x}}{{2\sin (\alpha  + \beta )x\sin (\alpha  - \beta )x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{[2\sin \alpha x\cos \beta x + 2\sin \alpha x\cos \alpha x]x}}{{2\sin (\alpha  + \beta )x\sin (\alpha  - \beta )x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin \alpha x[\cos \beta x + \cos \alpha x]x}}{{2\sin (\alpha  + \beta )x\sin (\alpha  - \beta )x}}$

$\left[ {\because \cos C - \cos D = 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{D - C}}{2}} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \alpha x\left[ {2\cos \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x\cos \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x} \right]x}}{{2\sin \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x\cos \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x \cdot 2\sin \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x\cos \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}$

$\left[ {\because \cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}\operatorname{and} \sin 2\theta  = 2\sin \theta \cos \theta } \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \alpha x \cdot x}}{{2\sin \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x\sin \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}$

$= \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin \alpha x}}{{\alpha x}} \cdot x \cdot (\alpha x)}}{{2\sin \dfrac{{\left( {\dfrac{{\alpha  + \beta }}{2}} \right)x}}{{\left( {\dfrac{{\alpha  + \beta }}{2}} \right)x}} \cdot \sin \dfrac{{\left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}{{\left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}} \cdot \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x \cdot \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}$

$= \dfrac{{\dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \alpha x}}{{\alpha x}} \cdot \alpha {x^2}}}{{\mathop {\lim }\limits_{x \to 0} \sin \dfrac{{\left( {\dfrac{{\alpha  + \beta }}{2}} \right)x}}{{\left( {\dfrac{{\alpha  + \beta }}{2}} \right)x}}\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}{{\left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}} \cdot \left( {\dfrac{{{\alpha ^2} - {\beta ^2}}}{4}} \right){x^2}}}$

$= \dfrac{1}{2} \cdot \dfrac{{\alpha  \cdot 4}}{{{\alpha ^2} - {\beta ^2}}}\left[ {\dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \alpha x}}{{\alpha x}}}}{{\mathop {\lim }\limits_{x \to 0} \sin \dfrac{{\left( {\dfrac{{\alpha  + \beta }}{2}} \right)x}}{{\left( {\dfrac{{\alpha  + \beta }}{2}} \right)x}}\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}{{\left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}}}} \right]$

$\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}x \to 0 \Rightarrow kx \to 0{\text{ }}} \right]$

$= \dfrac{1}{2} \cdot \dfrac{{4\alpha }}{{{\alpha ^2} - {\beta ^2}}}$

$= \dfrac{{2\alpha }}{{{\alpha ^2} - {\beta ^2}}}$

49. $\mathop{\lim }\limits_{x \rightarrow \dfrac{\pi}{4}} \dfrac{\tan ^{3} x-\tan x}{\cos x+\dfrac{\pi}{4}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\tan }^3}x - \tan x}}{{\cos \left( {x + \dfrac{\pi }{4}} \right)}}$

We know that,

$\dfrac{0}{0}{\text{ form}}$

$\because {a^2} - {b^2} = (a + b)(a - b)$

$\because \cos A \cdot \cos B - \sin A\sin B = \cos (A + B)$

Let,

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\tan }^3}x - \tan x}}{{\cos \left( {x + \dfrac{\pi }{4}} \right)}}$$\left[ {\dfrac{0}{0}{\text{ form}}} \right]$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\tan x\left( {{{\tan }^2}x - 1} \right)}}{{\cos \left( {x + \dfrac{\pi }{4}} \right)}} = \mathop {\lim }\limits_{x \to \pi /4} \tan x \cdot \mathop {\lim }\limits_{x \to \pi /4} \left( {\dfrac{{1 - {{\tan }^2}x}}{{\cos \left( {x + \dfrac{{g\pi }}{4}} \right)}}} \right)$

$\left[ {\because {a^2} - {b^2} = (a + b)(a - b)} \right]$

$=  - 1 \times \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{(1 + \tan x)(1 - \tan x)}}{{\cos \left( {x + \dfrac{\pi }{4}} \right)}}$

${=  - \mathop {\lim }\limits_{x \to \pi /4} (1 + \tan x)\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\cos x - \sin x}}{{\cos x \cdot \cos \left( {x + \dfrac{\pi }{4}} \right)}}}$

$=  - (1 + 1) \times \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sqrt 2 \left[ {\dfrac{1}{{\sqrt 2 }} \cdot \cos x - \dfrac{1}{{\sqrt 2 }} \cdot \sin x} \right]}}{{\cos x \cdot \cos \left( {x + \dfrac{\pi }{4}} \right)}}$

$=  - 2\sqrt 2 \mathop {\lim }\limits_{x \to \pi /4} \left[ {\dfrac{{\cos \dfrac{\pi }{4} \cdot \cos x - \sin \dfrac{\pi }{4} \cdot \sin x}}{{\cos x \cdot \cos \left( {x + \dfrac{\pi }{4}} \right)}}} \right]$

$\because \cos A \cdot \cos B - \sin A\sin B = \cos (A + B)$

$=  - 2\sqrt 2 \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\cos \left( {x + \dfrac{\pi }{4}} \right)}}{{\cos x \cdot \cos \left( {x + \dfrac{\pi }{4}} \right)}}$

$=  - 2\sqrt 2  \times \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}}$

$=  - 2\sqrt 2  \times \sqrt 2$

$=  - 4$

50. $\mathop{\lim }\limits _{x \rightarrow \pi} \dfrac{1-\sin \dfrac{x}{2}}{\cos \dfrac{x}{2} \cos \dfrac{x}{4}-\sin \dfrac{x}{4}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi } \dfrac{{1 - \sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}$

We know that,

$\because {\sin ^2}\theta  + {\cos ^2}\theta  = \sin 2\theta  = 2\sin \theta \cos \theta$

$\because {\cos ^2}2\theta  = {\cos ^2}\theta  - {\sin ^2}\theta$

$\because {a^2} - {b^2} = (a + b)(a - b]$

Let,

$\mathop {\lim }\limits_{x \to \pi } \dfrac{{1 - \sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{{{\cos }^2}\dfrac{x}{4} + {{\sin }^2}\dfrac{x}{4} - 2 \cdot \sin \dfrac{x}{4} \cdot \cos \dfrac{x}{4}}}{{\cos \dfrac{x}{2} \cdot \left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}\left[ {\because {{\sin }^2}\theta  + {{\cos }^2}\theta  = 1\sin 2\theta  = 2\sin \theta \cos \theta } \right]$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{{{\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}^2}}}{{\left( {{{\cos }^2}\dfrac{x}{4} - {{\sin }^2}\dfrac{x}{4}} \right)\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}\quad \left[ {\because {{\cos }^2}2\theta  = {{\cos }^2}\theta  - {{\sin }^2}\theta } \right]$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}{{\left( {\cos \dfrac{x}{4} + \sin \dfrac{x}{4}} \right)\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}\quad \left[ {\because {a^2} - {b^2} = (a + b)(a - b} \right]$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{1}{{\cos \dfrac{x}{4} + \sin \dfrac{x}{4}}}$

$= \dfrac{1}{{\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}}}$

$= \dfrac{{\sqrt 2 }}{2}$

$= \dfrac{1}{{\sqrt 2 }}$

51. Show that $\mathop{\lim }\limits_{x \rightarrow 4} \dfrac{|x-4|}{x-4}$ does not exists

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{|x - 4|}}{{x - 4}}$

Let,

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{|x - 4|}}{{x - 4}}$

${\text{LHL}} = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ - }}}{4}} \dfrac{{ - (x - 4)}}{{x - 4}}\quad [\because |x - 4| =  - (x - 4),x < 4]$

${\text{       }} =  - 1$

${\text{RHL}} = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{4}} \dfrac{{(x - 4)}}{{x - 4}}[\because |x - 4| = (x - 4),x > 4]$

${\text{        }} = 1$

$\therefore {\text{LHL}} \ne {\text{RHL}}$

So, limit does not exist

52. Let $f(x)=\dfrac{k \cos x}{\pi-2 x}, when x \neq \dfrac{\pi}{2} \& 3, x=\dfrac{\pi}{2}$ and if $\mathop{\lim }\limits_{x \rightarrow \dfrac{\pi}{2}}$ $f(x)=f\left(\dfrac{\pi}{2}\right)$, find the value of $k$.

Ans:

$\therefore {\text{LHL}} = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ - }}}{2}} \dfrac{{k\cos x}}{{\pi  - 2x}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{k\cos \left( {\dfrac{\pi }{2} - h} \right)}}{{\pi  - 2\left( {\dfrac{\pi }{2} - h} \right)}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{k\sin h}}{{\pi  - \pi  + 2h}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{\text{ksin}}h}}{{2h}}$

$= \dfrac{k}{2}\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin h}}{h}$

$= \dfrac{k}{2} \cdot 1$

$= \dfrac{k}{2}$

${\text{RHL}} = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{2}} \dfrac{{k\cos x}}{{\pi  - 2x}}$

$= \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}}  + \dfrac{{k\cos \left( {\dfrac{\pi }{2} + h} \right)}}{{\pi  - 2\left( {\dfrac{\pi }{2} + h} \right)}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{ - k\sin h}}{{\pi  - \pi  - 2h}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{k\sin h}}{{2h}}$

$= \dfrac{k}{2}\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin h}}{h}$

$= \dfrac{k}{2}$

${\text{ }}f\left( {\dfrac{\pi }{2}} \right) = 3$ (given)

It is given that,

$\mathop {\lim }\limits_{x \to \pi /2} f(x) = f\left( {\dfrac{\pi }{2}} \right)$

$\Rightarrow \dfrac{k}{2} = 3$

$\therefore k = 6$

53. Let $f(x)= \left\{\begin{align} &  x + 2, x \leq -1 \\ &  c{x^2}, x >  - 1 \end{align} \right.$, find ' $c$ ' if $\mathop{\lim }\limits_{x \rightarrow-1} f(x)$ exists.

Ans: $f(x)= \left\{\begin{align} &  x + 2, x \leq -1 \\ &  c{x^2}, x >  - 1 \end{align} \right.$

${\text{LHL}} = \mathop {\lim }\limits_{x \to  - {1^ - }} f(x)$

$= \mathop {\lim }\limits_{x \to  - {1^ - }} (x + 2)$

$= \mathop {\lim }\limits_{x \to  - {1^ - }} (x + 2)$

$= \mathop {\lim }\limits_{h \to 0} (1 - h) = 1$

${\text{RHL}} = \mathop {\lim }\limits_{x \to  - {1^ + }} f(x)$

$= \mathop {\lim }\limits_{x \to  - {1^ + }} c{x^2}$

$= \mathop {\lim }\limits_{h \to 0} c{( - 1 + h)^2}$

$= c$

If $\mathop {\lim }\limits_{x \to  - 1} f\left( x \right)$ exist, then $LHL = RHL$

$\therefore c = 1$

Objective Type Questions:

Choose the correct answer out of 4 options given against each Exercise 54 to 76 (M.C.Q).

54. $\mathop{\lim }\limits _{x \rightarrow \pi} \dfrac{\sin x}{x-\pi}$ is

(A) 1

(B) 2

(C) $-1$

(D) $-2$

Ans: 

Given: 

$\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin x}}{{x - \pi }}$

We know that,

$\because \sin \left( {\pi  - \,\theta } \right)\, = \,\sin \theta$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}\pi  - x \to 0 \Rightarrow x \to \pi$

Let,     

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin x}}{{x - \pi }}$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin \left( {\pi  - x} \right)}}{{ - \left( {\pi  - x} \right)}}{\text{  }}\left[ {{\text{ }}\because \sin \left( {\pi  - \theta } \right) = \sin \theta {\text{ }}} \right]$

$=  - \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin \left( {\pi  - x} \right)}}{{\left( {\pi  - x} \right)}}\left[ {{\text{ }}\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}\pi  - x \to 0 \Rightarrow x \to \pi {\text{ }}} \right]{\text{ }}$

$=  - 1$ 

Correct Option:  C 

55. $\mathop{\lim }\limits _{x \rightarrow 0} \dfrac{x^{2} \cos x}{1-\cos x}$ is

(A) 2

(B) $\dfrac{3}{2}$

(C) $\dfrac{-3}{2}$

(D) 1

Ans: Given: 

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\cos x}}{{1 - \cos x}}$

We know that,

$\because {\text{ }}1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\cos x}}{{1 - \cos x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\cos x}}{{2{{\sin }^2}\dfrac{x}{2}}}\quad \left[ {\because 1 - \cos x = 2{{\sin }^2}\dfrac{x}{2}} \right]$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {\dfrac{x}{2}} \right)}^2}}}{{{{\sin }^2}\dfrac{x}{2}}} \cdot \mathop {\lim }\limits_{x \to 0} \cos x$

$= 2 \cdot 1 = 2$ 

Correct Option:  A

56. $\mathop{\lim }\limits_{x \rightarrow 0} \dfrac{(1+x)^{n}-1}{x}$ is

(A) $n$

(B) 1

(C) $-n$

(D) 0

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {1 + x} \right)}^n} - 1}}{x}$

We know that,

$\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {1 + x} \right)}^n} - 1}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^n} - 1}}{{(1 + x) - 1}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^n} - {1^n}}}{{(1 + x) - 1}}$

$= \mathop {\lim }\limits_{(1 + x) \to 1} \dfrac{{{{(1 + x)}^n} - {1^n}}}{{(1 + x) - 1}}$

$= n \cdot {(1)^{n - 1}}$\[\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

$= n$

Correct Option:  A

57. $\mathop{\lim }\limits_{x \rightarrow 1} \dfrac{x^{m}-1}{x^{n}-1}$ is

(A) 1

(B) $\dfrac{m}{n}$

(C) $-\dfrac{m}{n}$

(D) $\dfrac{m^{2}}{n^{2}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^m} - 1}}{{{x^n} - 1}}$

We know that,

\[\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}\]

Then,

$=\lim _{x \rightarrow 1 \atop \rightarrow 1} \dfrac{x^{m}-1}{x^{n}-1}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{{x^m} - 1}}{{x - 1}}}}{{\dfrac{{{x^n} - 1}}{{x - 1}}}}$ 

$= \dfrac{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^m} - {1^m}}}{{x - 1}}}}{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^n} - {1^n}}}{{x - 1}}}}$

$=\lim _{x \rightarrow 1 \atop \rightarrow 1} \dfrac{x^{m}-1}{x^{n}-1}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{{x^m} - 1}}{{x - 1}}}}{{\dfrac{{{x^n} - 1}}{{x - 1}}}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^m} - {1^m}}}{{x - 1}}}}{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^n} - {1^n}}}{{x - 1}}}}$

$= \dfrac{{m{{(1)}^{m - 1}}}}{{n{{(1)}^{n - 1}}}}$

$= \dfrac{m}{n}\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$ 

Correct Option:  B

58. $\mathop{\lim }\limits_{x \rightarrow 0} \dfrac{1-\cos 4 \theta}{1-\cos 6 \theta}$ is

(A) $\dfrac{4}{9}$

(B) $\dfrac{1}{2}$

(C) $\dfrac{-1}{2}$

(D) $-1$

Ans: Given:

$\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{1 - \cos 4\theta }}{{1 - \cos 6\theta }}$

We know that,

$\because 1 - \cos 2\theta  = 2{\sin ^2}\theta$

\[\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}x \to 0 \Rightarrow kx \to 0\]

Let,

$\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{1 - \cos 4\theta }}{{1 - \cos 6\theta }}$

$= \mathop {\lim }\limits_{\theta  \to 0} \dfrac{{2{{\sin }^2}2\theta }}{{2{{\sin }^2}3\theta }}\quad \left[ {\because 1 - \cos 2\theta  = 2{{\sin }^2}\theta } \right]$

\[ = \dfrac{{\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{{{\sin }^2}2\theta }}{{{{(2\theta )}^2}}} \cdot {{(2\theta )}^2}}}{{\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{{{\sin }^2}3\theta }}{{{{(3\theta )}^2}}} \cdot {{(3\theta )}^2}}}{\text{ }}\]

\[ = \dfrac{4}{9} \cdot \dfrac{{\mathop {\lim }\limits_{\theta  \to 0} {{\left( {\dfrac{{\sin 2\theta }}{{2\theta }}} \right)}^2}}}{{\mathop {\lim }\limits_{\theta  \to 0} {{\left( {\dfrac{{\sin 3\theta }}{{3\theta }}} \right)}^2}}}\quad \left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]\]

$= \dfrac{4}{9}{\text{                            }}\left[ {x \to 0 \Rightarrow kx \to 0} \right]$

Correct Option:  A

59. $\mathop{\lim }\limits_{x \rightarrow 0} \dfrac{\operatorname{cosec} x-\cot x}{x}$ is

(A) $\dfrac{-1}{2}$

(B) 1

(C) $\dfrac{1}{2}$

(D) 1

Ans: Given:

\[\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{\operatorname{co} {\text{sec }}x - \cot x}}{x}{\text{ }}\]

We know that,

$\because \mathop {\lim }\limits_{\theta  \to 0} \dfrac{{\tan \theta }}{\theta } = 1$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\operatorname{cosec} x - \cot x}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{\sin x}} - \dfrac{{\cos x}}{{\sin x}}}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{x \cdot \sin x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{x \cdot 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan \dfrac{x}{2}}}{x}$${\text{  }}\left[ {\because \mathop {\lim }\limits_{\theta  \to 0} \dfrac{{\tan \theta }}{\theta } = 1} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan \dfrac{x}{2}}}{{\dfrac{x}{2}}} \cdot \dfrac{1}{2}$

$= \dfrac{1}{2}$

Correct Option:  C

60. $\mathop{\lim }\limits _{x \rightarrow 0} \dfrac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$ is

(A) 2

(B) 0

(C) 1

(D) $-1$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{\sqrt {x + 1}  - \sqrt {x - 1} }}$

We know that,

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{\sqrt {x + 1}  - \sqrt {1 - x} }}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{\sqrt {x + 1}  - \sqrt {1 - x} }} \cdot \dfrac{{\sqrt {x + 1}  + \sqrt {1 - x} }}{{\sqrt {x + 1}  + \sqrt {1 - x} }}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x(\sqrt {x + 1}  + \sqrt {1 - x} )}}{{(x + 1) - (1 - x)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x(\sqrt {x + 1}  + \sqrt {1 - x} )}}{{x + 1 - 1 + x}}$

$= \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}\mathop {\lim }\limits_{x \to 0} (\sqrt {x + 1}  + \sqrt {1 - x} )$

$= \dfrac{1}{2} \cdot 1 \cdot 2\,\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 1\quad$ 

Correct Option:  C

61. $\mathop{\lim }\limits_{x \rightarrow \dfrac{\pi}{4}} \dfrac{\sec ^{2} x-2}{\tan x-1}$ is

(A) 3

(B) 1

(C) 0

(D) $\sqrt{2}$

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{\text{se}}{{\text{c}}^2}{\text{ }}x - 2}}{{{\text{tan }}x - 1}}$

We know that,

\[\because \sec x = 1 + \tan x\]

\[\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Let,

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\sec }^2}x - 2}}{{\tan x - 1}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{1 + {{\tan }^2}x - 2}}{{\tan x - 1}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\tan }^2}x - 1}}{{\tan x - 1}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{(\tan x + 1)(\tan x - 1)}}{{(\tan x - 1)}}$

$= \mathop {\lim }\limits_{x \to \pi /4} (\tan x + 1)$

$= 2$

Correct Option:  D

62. $\mathop{\lim }\limits_{x \rightarrow 1} \dfrac{(\sqrt{x}-1)(2 x-3)}{2 x^{2}+x-3}$ is

(A) $\dfrac{1}{10}$

(B) $\dfrac{-1}{10}$

(C) 1

(D) None of these

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{(\sqrt x  - 1)(2x - 3)}}{{(2{x^2} + x - 3)}}$

We know that,

\[\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Let,

$\mathop {\lim }\limits_{x \to 1} \dfrac{{(\sqrt x  - 1)(2x - 3)}}{{2{x^2} + x - 3}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{(\sqrt x  - 1)(2x - 3)}}{{(2x + 3)(x - 1)}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{(\sqrt x  - 1)(2x - 3)}}{{(2x + 3)(\sqrt x  - 1)(\sqrt x  + 1)}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{2x - 3}}{{(2x + 3)(\sqrt x  + 1)}}$

$= \dfrac{{ - 1}}{{5 \times 2}}$

$= \dfrac{{ - 1}}{{10}}$

Correct Option:  B

63. If $f(x)=\dfrac{\sin [x]}{[x]}, [x] \neq 0$, $[x]=0$, where [.] denotes the greatest integer function, then $\mathop{\lim }\limits_{x \rightarrow 0} f(x)$ is equal to

(A) 1

(B) 0

(C) $-1$

(D) None of these

Ans:

Let,

$f\left( x \right) = \dfrac{{\sin \left[ x \right]}}{{\left[ x \right]}},{\text{  }}\left[ x \right] \ne 0$

$0,{\text{           }}\left[ x \right] = 0$ 

${\text{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} f(x)$

${\text{       }} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin [x]}}{{[x]}}$

${\text{       }} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin [0 - h]}}{{[0 - h]}}$

${\text{       }} = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \sin [ - h]}}{{[ - h]}} =  - 1$

${\text{RHL}} = \mathop {\lim }\limits_{x \to {0^ + }} f(x)$

${\text{        }} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin [x]}}{{[x]}}$

${\text{        }} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin [0 + h]}}{{[0 + h]}}$

${\text{        }} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin [h]}}{{[h]}} = 1$

\[{\text{LHL}} \ne {\text{RHL}}\]

Therefore, the limit does not exist.

Correct Option:  D

64. $\mathop{\lim }\limits _{x \rightarrow 0} \dfrac{|\sin x|}{x}$ is

(A) 1

(B) $-1$

(C) does not exist

(D) None of these

Ans: 

\[\therefore {\text{ }}LHL = \mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{{ - \sin x}}{x}} \right)\]

\[{\text{           }} =  - \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin x}}{x}\]

\[{\text{           }} =  - 1\]

\[{\text{   }}RHL = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin x}}{x}\]

\[{\text{           }} = 1\]

\[\because LHL \ne RHL\]

Thus, the limit does not exist.

Correct Option :  C

65. Let $f(x)= \left\{ \begin{align} & {x}^{2}-1,0<x<2 \\& 2x+3,2 \leq x<3  \\ \end{align} \right.$, the quadratic equation whose roots are $\mathop {\lim }\limits_{x \rightarrow 2^{-}} f(x)$ and $\mathop {\lim }\limits_{x \rightarrow 2^{+}} f(x)$ is

(A) $x^{2}-6 x+9=0$

(B) $x^{2}-7 x+8=0$

(C) $x^{2}-14 x+49=0$

(D) $x^{2}-10 x+21=0$

Ans: $\therefore \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 1} \right)$

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} \left[ {{{(2 - h)}^2} - 1} \right]$

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} \left( {4 + {h^2} - 4h - 1} \right)$

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} \left( {{h^2} - 4h + 3} \right)$

${\text{                    }} = 3$

${\text{and}}\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} (2x + 3)$

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} [2(2 + h) + 3]$

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} (4 + 2h + 3)$

${\text{                    }} = 7$

Thus, quadratic equation whose roots are $3{\text{ and }}7$

${x^2} - \left( {3 + 7} \right)x + 3 \times 7 = 0$

$i.e,{\text{ }}{x^2} - 10x + 21 = 0$

Correct Option :  D

66. $\mathop {\lim }\limits_{x \rightarrow 0} \dfrac{\tan 2 x-x}{3 x-\sin x}$ is

(A) 2

(B) $\dfrac{1}{2}$

(C) $\dfrac{-1}{2}$

(D) $\dfrac{1}{4}$

Ans: Correct Option :  B

Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 2x - x}}{{3x - \sin x}}$

We know that,

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1$

$x \to 0 \Rightarrow kx \to 0$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 2x - x}}{{3x - \sin x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left[ {\dfrac{{\tan 2x}}{x} - 1} \right]}}{{x\left[ {3 - \dfrac{{\sin x}}{x}} \right]}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 0} 2 \times \dfrac{{\tan 2x}}{{2x}} - 1}}{{3 - \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}}}$ 

${\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1}$

$\left[ {x \to 0 \Rightarrow kx \to 0} \right]$

$\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1} \right]$ 

$= \dfrac{{2 - 1}}{{3 - 1}}$

$= \dfrac{1}{2}$

67. Let $f(x)=x-[x] ; \in \mathbf{R}$, then $f^{\prime} \dfrac{1}{2}$ is

(A) $\dfrac{3}{2}$

(B) 1

(C) 0

(D) $-1$

Ans: $f\left( x \right) = x - \left[ x \right]$ (given)

$f'\left( x \right) = 1 + 0$

${\text{         }} = 1$ 

\[\left[ {\because \dfrac{d}{{dx}}x = 1{\text{ and }}\dfrac{d}{{dx}}\left[ x \right] = 0} \right]\]

Therefore, \[f'\left( {\dfrac{1}{2}} \right) = 1\]

Correct Option:  B

68. If $y=\sqrt{x}+\dfrac{1}{\sqrt{x}}$, then $\dfrac{d y}{d x}$ at $x=1$ is

(A) 1

(B) $\dfrac{1}{2}$

(C) $\dfrac{1}{\sqrt{2}}$

(D) 0

Ans: Given:

\[y = \sqrt x  + \dfrac{1}{{\sqrt x }}\]

We know that,

We will differentiate the given equation using chain rule

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

Then,

$y = \sqrt x  + \dfrac{1}{{\sqrt x }}$

${\text{now, }}\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt x }} - \dfrac{1}{{2{x^{3/2}}}}$

$\therefore {\text{ }}{\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 1}} = \dfrac{1}{2} - \dfrac{1}{2}$

${\text{                 }} = 0$

Correct Option:  D

69. If $f(x)=\dfrac{x-4}{2 \sqrt{x}}$, then $f^{\prime}(1)$ is

(A) $\dfrac{5}{4}$

(B) $\dfrac{4}{5}$

(C) 1

(D) 0

Ans: Given:

$f\left( x \right) = \dfrac{{x - 4}}{{2\sqrt x }}$

We know that, Differentiate  \[f\left( x \right)\]

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

At last putting \[1\] in \[f'\left( x \right)\] to get the final answer

Then,$f\left( x \right) = \dfrac{{x - 4}}{{2\sqrt x }}$

$f'\left( x \right) = \dfrac{{2\sqrt x  - (x - 4) \cdot 2 \cdot \dfrac{1}{{2\sqrt x }}}}{{4x}}$       (by quotient rule)

${\text{         }} = \dfrac{{2x - (x - 4)}}{{4{x^{3/2}}}}$

${\text{          }} = \dfrac{{2x - x + 4}}{{4{x^{3/2}}}}$

${\text{          }} = \dfrac{{x + 4}}{{4{x^{3/2}}}}$

$f'\left( 1 \right) = \dfrac{{1 + 4}}{{4 \times {{(1)}^{3/2}}}}$

${\text{           = }}\dfrac{5}{4}$

Correct Option :  A

70. If $y=\dfrac{1+\dfrac{1}{x^{2}}}{1-\dfrac{1}{x^{2}}}$, then $\dfrac{d y}{d x}$ is

(A) $\dfrac{-4 x}{\left(x^{2}-1\right)^{2}}$

(B) $\dfrac{-4 x}{x^{2}-1}$

(C) $\dfrac{1-x^{2}}{4 x}$

(D) $\dfrac{4 x}{x^{2}-1}$

Ans: Given:

$y = \dfrac{{1 + \dfrac{1}{{{x^2}}}}}{{1 - \dfrac{1}{{{x^2}}}}}$

We know that, We will differentiate the given equation using quotient rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

Then,

\[y = \dfrac{{1 + \dfrac{1}{{{x^2}}}}}{{1 - \dfrac{1}{{{x^2}}}}}\]

\[{\text{  }} = \dfrac{{{x^2} + 1}}{{{x^2} - 1}}\]

Differentiating \[y\] w.r.t  \[x\]

\[\dfrac{{dy}}{{dx}} = \dfrac{{\left( {{x^2} - 1} \right)2x - \left( {{x^2} + 1} \right)(2x)}}{{{{\left( {{x^2} - 1} \right)}^2}}}{\text{ }}\left[ {{\text{By Quotient rule}}} \right]\]

\[ = \dfrac{{2x\left( {{x^2} - 1 - {x^2} - 1} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}\]

\[ = \dfrac{{2x( - 2)}}{{{{\left( {{x^2} - 1} \right)}^2}}}\]

\[ = \dfrac{{ - 4x}}{{{{\left( {{x^2} - 1} \right)}^2}}}\]

Correct Option:  A

71. If $y=\dfrac{\sin x+\cos x}{\sin x-\cos x} \text {, then } \dfrac{d y}{d x} \text { at } x=0$ is

(A) $-2$

(B) 0

(C) $\dfrac{1}{2}$

(D) does not exist

Ans: Given, 

\[y = \dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}\]

Now, Differentiating \[y\] w.r.t  \[x\]

\[\dfrac{{dy}}{{dx}} = \dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}\]

\[\dfrac{{dy}}{{dx}} = \dfrac{{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}}{{{{(\sin x - \cos x)}^2}}}{\text{ }}\left[ {{\text{by quotient rule}}} \right]\]$= \dfrac{{ - {{(\sin x - \cos x)}^2} - {{(\sin x + \cos x)}^2}}}{{{{(\sin x - \cos x)}^2}}}$

$= \dfrac{{ - \left[ {{{(\sin x - \cos x)}^2} + {{(\sin x + \cos x)}^2}} \right]}}{{{{(\sin x - \cos x)}^2}}}$

$= \dfrac{{ - \left[ {{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x + {{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right]}}{{{{(\sin x - \cos x)}^2}}}$

$= \dfrac{{ - 2}}{{{{(\sin x - \cos x)}^2}}}$

\[\therefore {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} =  - 2\]

Correct Option:  A

72. If $y=\dfrac{\sin (x+9)}{\cos x}$ then $\dfrac{d y}{d x}$ at $x=0$ is

(A) $\cos 9$

(B) $\sin 9$

(C) 0

(D) 1

Ans: Given, $y = \dfrac{{\sin (x + 9)}}{{\cos x}}$

 Differentiating \[y\] w.r.t  \[x\]

$\dfrac{{dy}}{{dx}} = \dfrac{{\cos x\cos (x + 9) - \sin (x + 9)( - \sin x)}}{{{{(\cos x)}^2}}}\left[ {{\text{by quotient rule}}} \right]$

$= \dfrac{{\cos x\cos (x + 9) + \sin x\sin (x + 9)}}{{{{\cos }^2}x}}\left[ {\because \cos x\cos y + \sin x\sin y = \cos \left( {x - y} \right)} \right]$

$= \dfrac{{\cos \left( {x + 9 - x} \right)}}{{{{\cos }^2}x}}$

${\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = \dfrac{{\cos 9}}{1}$

$= \cos 9$

Correct Option:  A

73. If $f(x)=1+x+\dfrac{x^{2}}{2}+\ldots+ \dfrac{x^{100}}{100}$, then $f^{\prime}(1)$ is equal to

(A) $\dfrac{1}{100}$

(B) 100

(C) does not exist

(D) 0

Ans: Given:

$f\left( x \right) = 1 + x + \dfrac{{{x^2}}}{2} + ... + \dfrac{{{x^{100}}}}{{100}}$

We know that,Differentiating the \[f\left( x \right)\] w.r.t  \[x\]

Then, putting \[x = 1\] in the solved equation

Then,

\[{\text{          }}f\left( x \right) = 1 + x + \dfrac{{{x^2}}}{2} + ... + \dfrac{{{x^{100}}}}{{100}}\]

 Differentiating the equation w.r.t  \[x\]

\[{\text{         }}f\left( x \right)' = 0 + 1 + 2 \times \dfrac{x}{2} + ... + \dfrac{{{x^{100}}}}{{100}}\]

\[\therefore {\text{      }}f\left( x \right)' = 1 + x + {x^2} + ... + {x^{99}}\]

Now, putting \[x = 1\]in the solved equation

\[{\text{Now, }}f\left( x \right)' = 1 + 1 + 1 + ... + 1\left( {100{\text{ times}}} \right)\]

\[{\text{                  }} = 100\]

Correct Option :  B

74. If $f(x)=\dfrac{x^{n}-a^{n}}{x-a}$ for some constant ' $a$ ', then $f^{\prime}(a)$ is

(A) 1

(B) 0

(C) does not exist

(D) $\dfrac{1}{2}$

Ans: Given:

\[{\text{ }}f\left( x \right) = \dfrac{{{x^n} - {a^n}}}{{x - a}}\]

We know that, Differentiating the given equation by quotient rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

Putting  \[a\] in the solved equation to get the final answer

Then,

\[{\text{ }}f\left( x \right) = \dfrac{{{x^n} - {a^n}}}{{x - a}}\]

Differentiating the given equation w.r.t  \[x\]

$\therefore {\text{ }}f'(x) = \dfrac{{(x - a)n{x^{n - 1}} - \left( {{x^n} - {a^n}} \right)(1)}}{{{{(x - a)}^2}}}{\text{ }}\left[ {{\text{by quotient rule}}} \right]$

$\Rightarrow f'(x) = \dfrac{{n{x^{n - 1}}(x - a) - {x^n} + {a^n}}}{{{{(x - a)}^2}}}$

${\text{Now, }}f'(a) = \dfrac{{n{a^{n - 1}}(0) - {a^n} + {a^n}}}{{{{(x - a)}^2}}}$

$\Rightarrow {\text{      }}{f^\prime }(a) = \dfrac{0}{0}$

So, \[f'\left( a \right)\] does not exist,

Since, \[f\left( x \right)\] is not defined at \[x = a\].

Hence, \[f'\left( x \right)\] at \[x = a\] does not exist.

Correct Option :  D

75. If $f(x)=x^{100}+x^{99}+\ldots+x+1$, then $f^{\prime}(1)$ is equal to

(A) 5050

(B) 5049

(C) 5051

(D) 50051

Ans: Given:

\[{\text{ }}f\left( x \right) = {x^{100}} + {x^{99}} + ... + x + 1\]

We know that,Differentiating $f\left( x \right)$ w.r.t $x$

Then using this formula to find sum

${S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\}$

Then, $f(x) = {x^{100}} + {x^{99}} +  \ldots  + x + 1$

Differentiating w.r.t $x$

$\therefore f'(x) = 100{x^{99}} + 99{x^{98}} +  \ldots  + 1 + 0$

$= 100{x^{99}} + 99{x^{98}} +  \ldots  + 1$

${\text{Now, }}f'(1) = 100 + 99 +  \ldots  + 1$

${\text{                 }} = \dfrac{{100}}{2}[2 \times 100 + (100 - 1)( - 1)]\,\left[ {\because {S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\} } \right]$

${\text{                 }} = 50[200 - 99]\quad$

${\text{                 }} = 50 \times 101$

${\text{                 }} = 5050$

Correct Option:  A

76. If $f(x)=1-x+x^{2}-x^{3} \ldots-x^{99}+x^{100}$, then $f^{\prime}(1)$ is equal to

(A) 150

(B) $-50$

(C) $-150$

(D) 50

Ans: Given:

$f\left( x \right) = 1 - x + {x^2} - {x^3} + ... - {x^{99}} + {x^{100}}$

We know that, Differentiating $f\left( x \right)$ w.r.t $x$

Then using this formula to find sum

${S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\}$

Then,

$f\left( x \right) = 1 - x + {x^2} - {x^3} + ... - {x^{99}} + {x^{100}}$

Differentiating above equation w.r.t $x$

$f'(x) = 0 - 1 + 2x - 3{x^2} +  \ldots  - 99{x^{98}} + 100{x^{99}}$

$=  - 1 + 2x - 3{x^2} +  \ldots  - 99{x^{98}} + 100{x^{99}}{f^\prime }(1)$

$=  - 1 + 2 - 3 +  \ldots  - 99 + 100$

$\therefore f'\left( x \right) = ( - 1 - 3 - 5 -  \ldots  - 99) + (2 + 4 +  \ldots  + 100)$

$=  - \dfrac{{50}}{2}[2 \times 1 + (50 - 1)2] + \dfrac{{50}}{2}[2 \times 2 + (50 - 1)2]\,{\text{ }}\left[ {{S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\} } \right]$

$=  - 25[2 + 49 \times 2] + 25[4 + 49 \times 2]$

$=  - 25(2 + 98) + 25(4 + 98)$

$=  - 25 \times 100 + 25 \times 102$

$=  - 2500 + 2550$

$= 50$

Correct Option :  A

Fill in the blanks in Exercises 77 to 80: 

77. If $f(x)=\dfrac{\tan x}{x-\pi}$, then $\mathop {\lim }\limits _{x \rightarrow \pi} f(x)=$

Ans: Given:

\[f\left( x \right) = \dfrac{{\tan x}}{{x - \pi }}\]

We know that,

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1{\text{ }}$

Then,

\[f\left( x \right) = \dfrac{{\tan x}}{{x - \pi }}\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi } f\left( x \right) = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\tan x}}{{x - \pi }}\]

\[{\text{                  }} = \mathop {\lim }\limits_{\pi  - x \to 0} \dfrac{{ - \tan \left( {\pi  - x} \right)}}{{x - \pi }}\]$\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1{\text{ }}} \right]$

\[{\text{                  }} = 1\]

78. $\mathop{\lim}\limits _{x \rightarrow 0} \sin m x \cot \dfrac{x}{\sqrt{3}}=2$, then $m=$

Ans: Given:

\[\mathop {\lim }\limits_{x \to 0} \left( {\sin mx{\text{cot}}\dfrac{x}{{\sqrt 3 }}} \right) = 2\]

We know that,

\[\because {\text{cot}}\dfrac{x}{{\sqrt 3 }} = \dfrac{1}{{\tan \dfrac{x}{{\sqrt 3 }}}}\]

Then,

\[\mathop {\lim }\limits_{x \to 0} \left( {\sin mx{\text{cot}}\dfrac{x}{{\sqrt 3 }}} \right) = 2\] (given)

Multiplying $\dfrac{{mx}}{{mx}}$ in L.H.S  $\left[ {\because \dfrac{{mx}}{{mx}} = 1} \right]$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin mx}}{{mx}} \cdot mx \cdot \dfrac{1}{{\tan \dfrac{x}{{\sqrt 3 }}}} = 2$\[\left[ {\because {\text{cot}}\dfrac{x}{{\sqrt 3 }} = \dfrac{1}{{\tan \dfrac{x}{{\sqrt 3 }}}}} \right]\]

Multiplying and dividing \[\dfrac{x}{{\sqrt 3 }}\]in numerator and denominator in L.H.S

$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin mx}}{{mx}} \cdot mx \cdot \dfrac{{\dfrac{x}{{\sqrt 3 }}}}{{\tan \dfrac{x}{{\sqrt 3 }}}} \cdot \dfrac{1}{{\dfrac{x}{{\sqrt 3 }}}} = 2$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin mx}}{{mx}} \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{x}{{\sqrt 3 }}}}{{\tan \dfrac{x}{{\sqrt 3 }}}} \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{{mx}}{{\dfrac{x}{{\sqrt 3 }}}} = 2$

$\Rightarrow \sqrt 3 x = 2$

$\therefore m = \dfrac{{2\sqrt 3 }}{3}$

79. if $y=1+\dfrac{x}{1 !}+\dfrac{x^{2}}{2 !}+\dfrac{x^{3}}{3 !}+\ldots$, then $\dfrac{d y}{d x}=$

Ans: Given:

$y = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ...$

Then,

$y = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} +  \ldots$ (given)

$\therefore \dfrac{{dy}}{{dx}} = 0 + 1 + \dfrac{{2x}}{2} + \dfrac{{3{x^2}}}{6} + \dfrac{{4{x^3}}}{{4!}}$

${\text{       }} = 1 + x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{6} +  \ldots$

${\text{       }} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} +  \ldots$

${\text{        }} = y$

80. $\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{x}{{\left[ x \right]}}=$

Ans: $\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{x}{{\left[ x \right]}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {3 + h} \right)}}{{\left[ {3 + h} \right]}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {3 + h} \right)}}{{\left[ 3 \right]}}$

$= 1$ 

Ncert Exemplar Solutions For Class 11 Mathematics Chapter 13 Limits And Derivatives

The NCERT Exemplar solutions for class 11 provide a comprehensive solution for all the questions in the NCERT Exemplar textbook to the students of class 11 maths. To perform well in the board examinations, the students can refer to these solutions, which will help them increase their level of confidence, as the concepts are clearly structured and explained by the subject experts at Vedantu. The solutions here are prepared and reviewed by the field experts and are revised according to the latest NCERT syllabus and the guidelines of the CBSE board. The NCERT Exemplar books are widely used by the students who want to excel in the board exams, as it provides them with a vast number of questions to solve and practice. With the help of the NCERT Exemplar Solutions, the class 11 students will be able to solve the complex problems in each exercise of the textbook.

Chapter 13 Limits and Derivatives of the NCERT Exemplar Solutions for Class 11 Maths is a good study resource. The chapter Limits and Derivatives explains the limits of a function. We at Vedantu have provided the answers to the NCERT Exemplar Solutions in simple PDF format for the students, which they can easily download from Vedantu’s website. The solutions will help the students to clarify their doubts and provide them with a strong foundation for every concept. 

Important Topics of Chapter-13 Limits and Derivatives

The important topics here include the following:

• Limits of a function

• Some properties of limits

• Limits of polynomials and rational functions

• Limits of trigonometric functions

• Derivatives

• Algebra of derivatives of functions