NCERT Solutions for Class 11 Maths Chapter 12 - Limits and Derivatives Exercise 12.1

Exercise 12.1

1. Evaluate the Given limit.$\underset{x\to 3}{\mathop{\lim }}\,x+3$ 

Ans. Given,

 $\underset{x\to 3}{\mathop{\lim }}\,x+3$

=$3+3$

=$6$

2. Evaluate the Given limit.$\underset{x\to \pi }{\mathop{\lim }}\,\left( x-\frac{22}{7} \right)$ 

Ans. Given,

$\underset{x\to \pi }{\mathop{\lim }}\,\left( x-\frac{22}{7} \right)$

=\[\left( \pi -\frac{22}{7} \right)\]

3. Evaluate the Given limit. $\underset{r\to 1}{\mathop{\lim }}\,\pi {{r}^{2}}$

Ans. Given,

$\underset{r\to 1}{\mathop{\lim }}\,\pi {{r}^{2}}$

=\[\pi ({{1}^{2}})\]

=\[\pi \]

4. Evaluate the Given limit.\[\underset{x\to 1}{\mathop{\lim }}\,\frac{4x+3}{x-2}\]

Ans. Given,

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{4x+3}{x-2}\]

=$\frac{4(4)+3}{4-2}$ 

=$\frac{16+3}{2}$ 

=$\frac{19}{2}$ 

5. Evaluate the Given limit. \[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{10}}+{{x}^{5}}+1}{x-1}\]

Ans. Given,

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{10}}+{{x}^{5}}+1}{x-1}\]

=$\frac{{{(-1)}^{10}}+{{(-1)}^{5}}+1}{-1-1}$ 

=$\frac{1-1+1}{-2}$ 

=$-\frac{1}{2}$ 

6. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}\]

Put $x+1=y$ So,

$y\to 1$ as $x\to 0$ 

Than, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}\]

=\[\underset{x\to 1}{\mathop{\lim }}\,\frac{(y)_{{}}^{5}-1}{y-1}\]

Using \[\left[ \underset{x\to a}{\mathop{\lim }}\,\frac{x_{{}}^{n}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]\]

=${{5.1}^{5-1}}$

=$5$

7. Evaluate the Given limit. \[\underset{x\to 2}{\mathop{\lim }}\,\frac{3{{x}^{2}}-x-10}{{{x}^{2}}-4}\]

Ans. At $x=2$

The value of the given rational function takes the form \[\frac{0}{0}\] 

=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{3{{x}^{2}}-x-10}{{{x}^{2}}-4}\]

=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)(3x+5)}{(x-2)(x+2)}\]

=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{3x+5}{x+2}\]

=\[\frac{3(2)+5}{2+2}\]

=$\frac{11}{4}$

8. Evaluate the Given limit. \[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{4}}-81}{2{{x}^{2}}-5x-3}\]

Ans. At $x=2$

The value of the given rational function takes the form \[\frac{0}{0}\] 

=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{4}}-81}{2{{x}^{2}}-5x-3}\]

=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)(x+3)({{x}^{2}}+9)}{(x-3)(2x+1)}\]

=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{(x+3)({{x}^{2}}+9)}{(2x+1)}\]

=\[\frac{(3+3)({{3}^{2}}+9)}{(2(3)+1)}\]

=\[\frac{6\times 18}{7}\]

=\[\frac{108}{7}\]

9. Evaluate the Given limit.\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+b}{cx+1}\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+b}{cx+1}\]

=\[\frac{a(0)+b}{c(0)+1}\]

=\[b\]

10. Evaluate the Given limit.\[\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{{{z}^{\frac{1}{6}}}-1}\]

Ans. At $z=1$

The value of the given rational function takes the form \[\frac{0}{0}\] 

Put ${{z}^{\frac{1}{6}}}=x$ So,

$z\to 1$ as $x\to 1$ 

Than,\[\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{{{z}^{\frac{1}{6}}}-1}\]

=\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{{{x}^{{}}}-1}\]

Using \[\left[ \underset{x\to a}{\mathop{\lim }}\,\frac{x_{{}}^{n}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]\]

=${{2.1}^{2-1}}$

=$2$

 11. Evaluate the Given limit. \[\underset{x\to 1}{\mathop{\lim }}\,\frac{a{{x}^{2}}+bx+c}{c{{x}^{2}}+bx+a}\],\[a+b+c\ne 0\]

Ans. Given,

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{a{{x}^{2}}+bx+c}{c{{x}^{2}}+bx+a}\]

=\[\frac{a{{(1)}^{2}}+b(1)+c}{c{{(1)}^{2}}+b(1)+a}\]

=\[\frac{a+b+c}{c+b+a}\]

=$1$                    \[(a+b+c\ne 0)\]

12. Evaluate the Given limit. \[\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\]

Ans. Given,

\[\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\]

At $x=-2$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\]=\[\underset{x\to -2}{\mathop{\lim }}\,\frac{\left( \frac{2+x}{2x} \right)}{x+2}\]

=\[\underset{x\to -2}{\mathop{\lim }}\,\frac{1}{2x}\]

=\[\frac{1}{2(-2)}\]

=\[-\frac{1}{4}\]

13. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}\]

At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}\times \frac{ax}{ax}\]

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{bx} \right)\times \frac{a}{b}\]

=\[\frac{a}{b}\underset{ax\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{bx} \right)\]

$x\to 0\Rightarrow ax\to 0$ 

=\[\frac{a}{b}\times 1\]\[\]         $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right) \right]$ 

=\[\frac{a}{b}\]

14. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx},a,b\ne 0\]

Ans.Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx},a,b\ne 0\]

At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{\sin ax}{ax} \right)\times ax}{\left( \frac{\sin ax}{ax} \right)\times bx}\]

=\[\frac{a}{b}\times \frac{\underset{ax\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{ax} \right)}{\underset{bx\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{ax} \right)}\]         $\left[ \begin{align} & x\to 0\Rightarrow ax\to 0 \\ & x\to 0\Rightarrow bx\to 0 \\ \end{align} \right]$

=\[\frac{a}{b}\times \frac{1}{1}\]                   $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

=\[\frac{a}{b}\]

15. Evaluate the Given limit. \[\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}\]

Ans. Given,

\[\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}\]

$\left[ x\to \pi \Rightarrow (\pi -x)\to 0 \right]$

\[\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}=\frac{1}{\pi }\underset{(\pi -x)\to 0}{\mathop{\lim }}\,\frac{\sin (\pi -x)}{(\pi -x)}\]

$=\frac{1}{\pi }\times 1$                      $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

$=\frac{1}{\pi }$

16. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{cosx}{\pi -x}\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\operatorname{cosx}}{\pi -x}\]

=\[\frac{\cos 0}{\pi -0}\]

=\[\frac{1}{\pi }\]

17. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}\]

At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-2{{\sin }^{2}}x-1}{1-2{{\sin }^{2}}\frac{x}{2}-1}\]

=\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x}{{{\sin }^{2}}\frac{x}{2}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)\times {{x}^{2}}}{\left( \frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)\times \frac{{{x}^{2}}}{4}}\]

=\[4\frac{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)}\]

=\[4\frac{{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)}^{2}}^{{}}}{{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)}^{2}}}\]

$\left[ x\to 0\Rightarrow \frac{x}{2}\to 0 \right]$

=\[4\frac{{{1}^{2}}^{{}}}{{{1}^{2}}}\]         $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

=\[4\]   

18. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}\]

Ans.Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}\]

At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}=\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\frac{x(a+\cos x)}{\sin x}\]

\[\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x}{\sin x} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right)\]

=\[\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin x}{x} \right)} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right)\]

=\[\frac{1}{b}\times \left( a+\cos 0 \right)\]               $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

  =\[\frac{a+1}{b}\]         

19. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,x\sec x\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,x\sec x\]

  =\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\cos x}\]        

  =\[\frac{0}{\cos 0}\]

=\[\frac{0}{1}\]

=\[0\]

20. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax+bx}{ax+\sin bx}\]\[a,b,a+b\ne 0\]

Ans. At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax+bx}{ax+\sin bx}\]

  =\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{\sin ax}{ax} \right)ax+bx}{ax+bx\left( \frac{\sin bx}{bx} \right)}\]        

  =\[\frac{\left( \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin ax}{ax} \right)\times \underset{x\to 0}{\mathop{\lim }}\,(ax)+\underset{x\to 0}{\mathop{\lim }}\,(bx)}{\underset{x\to 0}{\mathop{\lim }}\,ax+\underset{x\to 0}{\mathop{\lim }}\,bx\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin bx}{bx} \right)}\]

\[\left[ x\to \pi \Rightarrow ax\to 0 \right]\] and \[\left[ bx\to 0 \right]\]

  =\[\frac{\underset{x\to 0}{\mathop{\lim }}\,(ax)+\underset{x\to 0}{\mathop{\lim }}\,(bx)}{\underset{x\to 0}{\mathop{\lim }}\,ax+\underset{x\to 0}{\mathop{\lim }}\,bx}\]         $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

  =\[\frac{\underset{x\to 0}{\mathop{\lim }}\,(ax+bx)}{\underset{x\to 0}{\mathop{\lim }}\,(ax+bx)}\]   

  =\[\underset{x\to 0}{\mathop{\lim }}\,(1)\]         

  =\[1\]         

21. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,(\operatorname{cosecx}-cotx)\]

Ans. At $x=0$

The value of the given rational function takes the form \[\infty \to \infty \] 

\[\underset{x\to 0}{\mathop{\lim }}\,(\operatorname{cosecx}-cotx)\]

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{\sin x}-\frac{\cos x}{\sin x} \right)\]         

  =\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos x}{\sin x} \right)\]       

  =\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{1-\cos x}{x} \right)}{\left( \frac{\sin x}{x} \right)}\]         

  =\[\frac{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos x}{x} \right)}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin x}{x} \right)}\]   

$\left[ \underset{y\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{x}=0 \right]$ and $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

  =\[\frac{0}{1}\]        

  =\[0\]               

22. Evaluate the Given limit. \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}\]

Ans. Given, 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}\]

At $x=\frac{\pi }{2}$

The value of the given rational function takes the form \[\frac{0}{0}\] 

Put $x-\frac{\pi }{2}=y$ 

So,\[\left[ x\to \frac{\pi }{2},y\to 0 \right]\]

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}=\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan 2\left( y+\frac{\pi }{2} \right)}{y}\]

  =\[\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan (\pi +2y)}{y}\]       

  =\[\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan 2y}{y}\]    \[\left[ \tan (\pi +2y)=\tan 2y \right]\]

  =\[\underset{y\to 0}{\mathop{\lim }}\,\frac{\sin 2y}{y\cos 2y}\]

  =\[\underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\sin 2y}{2y}\times \frac{2}{\cos 2y} \right)\]

  =\[\underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\sin 2y}{2y} \right)\times \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{2}{\cos 2y} \right)\]  \[\left[ y\to 0\Rightarrow 2y\to 0 \right]\]

  =\[1\times \frac{2}{\cos 0}\]    \[\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]\]

  =\[1\times \frac{2}{1}\]

  =\[2\]

23. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$and $\underset{x\to 1}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & 2x+3 \\ & 3(x+1) \\ \end{align} \right.\]  $\begin{align} & x\le 0 \\ & x>0 \\ \end{align}$

Ans. Given.

\[f(x)=\left\{ \begin{align} & 2x+3 \\ & 3(x+1) \\ \end{align} \right.\]    $\begin{align} & x\le 0 \\ & x>0 \\ \end{align}$

\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2x+3 \right]\] 

  =\[2(0)+3=3\]

\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\left[ 3x+1 \right]\] 

  =\[3(0+1)=3\]

Therefore,

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,f(x)=3$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,(x+1)=3(1+1)=6$

Therefore,

$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(x)=6$

24. Find $\underset{x\to 1}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & {{x}^{2}}-1 \\ & -x-1 \\ \end{align} \right.\]   $\begin{align} & x\le 1 \\ & x>1 \\ \end{align}$

Ans. Given, 

\[f(x)=\left\{ \begin{align} & {{x}^{2}}-1 \\ & -x-1 \\ \end{align} \right.\]     $$\begin{align} & x\le 1 \\ & x>1 \\ \end{align}$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,({{x}^{2}}-1)={{1}^{2}}-1=1-1=0$

So, it is observed that 

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)$ 

Hence, $\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.

25. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & \frac{\left| x \right|}{x} \\ & 0 \\ \end{align} \right.\]    $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

Ans. Given

\[f(x)=\left\{ \begin{align} & \frac{\left| x \right|}{x} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{\left| x \right|}{x} \right]$

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{-x}{x} \right)\]    when x is negative, $\left| x \right|=-x$ 

=\[\underset{x\to 0}{\mathop{\lim }}\,(-1)\]

=\[-1\]

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \frac{\left| x \right|}{x} \right]$     

=$\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x}{x} \right)$

 when x is positive, $\left| x \right|=x$ 

=$\underset{x\to 0}{\mathop{\lim }}\,(1)$

=\[1\]

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$

Hence, $\underset{x\to {{0}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.

26. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & \frac{x}{\left| x \right|} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

Ans. Given

\[f(x)=\left\{ \begin{align} & \frac{x}{\left| x \right|} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$ 

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{x}{\left| x \right|} \right]$

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{-x}{x} \right)\]    when x is negative, $\left| x \right|=-x$ 

=\[\underset{x\to 0}{\mathop{\lim }}\,(-1)\]

=\[-1\]

 $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \frac{x}{\left| x \right|} \right]$   
=$\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x}{x} \right)$

 when x is positive, $\left| x \right|=x$ 

=\[\underset{x\to 0}{\mathop{\lim }}\,(1)\]

=\[1\]

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$

Hence, $\underset{x\to {{0}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.

27. Find $\underset{x\to 5}{\mathop{\lim }}\,f(x),$ where $f(x)=\left| x \right|-5$ 

Ans. Given,

 $f(x)=\left| x \right|-5$

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,(\left| x \right|-5)$ 

= $\underset{x\to 5}{\mathop{\lim }}\,(x-5)$    when x is positive, $\left| x \right|=x$ 

= $5-5$ 

= $0$ 

$\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,(\left| x \right|-5)$ 

= $\underset{x\to 5}{\mathop{\lim }}\,(x-5)$ 

when x is positive, $\left| x \right|=x$ 

= $5-5$ 

= $0$ 

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f(x)$

Hence, $\underset{x\to 5}{\mathop{\lim }}\,f(x)=0$ 

28. Suppose $f(x)=\left\{ \begin{align} & a+bx \\ & 4 \\ & b-ax \\ \end{align} \right.$   $\begin{align} & x<0 \\ & x=1 \\ & x>1 \\ \end{align}$and if $\underset{x\to 1}{\mathop{\lim }}\,f(x)=f(1)$what are possible values of a and b?

Ans. The given function is

$f(x)=\left\{ \begin{align} & a+bx \\ & 4 \\ & b-ax \\ \end{align} \right.$ $\begin{align} & x<0 \\ & x=1 \\ & x>1 \\ \end{align}$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(ax+bx)=a+b$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(b-ax)=b-a$

$f(1)=4$

Given

$\underset{x\to 1}{\mathop{\lim }}\,f(x)=f(1)$ 

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,f(x)=f(1)$

$a+b=4$and $b-a=4$

On solving,we get

$a=0$and $b=4$

29. Let a1,a2,……an be fixed real number and define a fuction

$f(x)=(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})$ 

What is $\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,f(x)$ ? For some a$\ne $ a1, a2,…..,an. Compute $\underset{x\to a}{\mathop{\lim }}\,f(x)$.

Ans. Given,

$f(x)=(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})$

$\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,[(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})]$

=\[({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})=0\]

Therefore,

$\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,f(x)=0$

Now, $\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{x\to a}{\mathop{\lim }}\,[(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})]$

=\[({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})\]

Therefore,

$\underset{x\to a}{\mathop{\lim }}\,f(x)=({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})$

30. If \[f(x)=\left\{ \begin{align} & \left| x \right|+1 \\ & 0 \\ & \left| x \right|-1 \\ \end{align} \right.\] $\begin{align} & x<0 \\ & x=0 \\ & x>1 \\ \end{align}$

For what value (s) of does $\underset{x\to a}{\mathop{\lim }}\,f(x)$  exists? 

Ans. Given,

 \[f(x)=\left\{ \begin{align} & \left| x \right|+1 \\ & 0 \\ & \left| x \right|-1 \\ \end{align} \right.\] $\begin{align} & x<0 \\ & x=0 \\ & x>1 \\ \end{align}$ 

When $a=0$ 

=\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( \left| x \right|+1 \right)\]

=\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -x+1 \right)\]     when x is negative, $\left| x \right|=-x$ 

=$0+1$

=$1$ 

=\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \left| x \right|+1 \right)\]

=\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( x-1 \right)\]     when x is positive, $\left| x \right|=x$ 

=$0-1$

=$-1$ 

Here, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$ 

$\underset{x\to 0}{\mathop{\lim }}\,f(x)$ does not exist.

When\[\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,(\left| x \right|+1)\] $a<0$

\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(\left| x \right|+1)\]

=\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(-x+1)\]           $[x<a<0\Rightarrow \left| x \right|=-x]$

=\[-a+1\]

$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=-a+1$ 

Thus, limit exists at $x=a$, where \[a<0\]

When \[a>0\]

$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(\left| x \right|+1)$

=$\underset{x\to a}{\mathop{\lim }}\,(-x-1)$                         $[0<x<a\Rightarrow \left| x \right|=x]$

=$-a-1$

$\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,(\left| x \right|+1)$

=$\underset{x\to a}{\mathop{\lim }}\,(-x-1)$                         $[0<x<a\Rightarrow \left| x \right|=x]$

=$a-1$

$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=a-1$

Thus, limit exists at $x=a$, where \[a>0\]

Thus $\underset{x\to a}{\mathop{\lim }}\,f(x)$ exists for all $a\ne 0$$\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi $.

31. If function f(x) satisfies, \[\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi \], evaluate $\underset{x\to 1}{\mathop{\lim }}\,f(x)$

Ans. Given, 

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi \]

= $\frac{\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)}{\underset{x\to 1}{\mathop{\lim }}\,({{x}^{2}}-1)}=\pi $

= $\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=\pi \underset{x\to 1}{\mathop{\lim }}\,({{x}^{2}}-1)$

=$\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=\pi ({{1}^{2}}-1)$

=$\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=0$

=$\underset{x\to 1}{\mathop{\lim }}\,f(x)-\underset{x\to 1}{\mathop{\lim }}\,2=0$

=$\underset{x\to 1}{\mathop{\lim }}\,f(x)-2=0$

$\underset{x\to 1}{\mathop{\lim }}\,f(x)=2$

32. If $f(x)=\left\{ \begin{align} & m{{x}^{2}}+n \\ & nx+m \\ & n{{x}^{3}}+m \\ \end{align} \right.$ $\begin{align} & x<0 \\ & 0\le x\le 1 \\ & x>1 \\ \end{align}$

For what integers m and n does $\underset{x\to 0}{\mathop{\lim }}\,f(x)$ and $\underset{x\to 1}{\mathop{\lim }}\,f(x)$ exists?

Ans. Given,  

$f(x)=\left\{ \begin{align} & m{{x}^{2}}+n \\ & nx+m \\ & n{{x}^{3}}+m \\ \end{align} \right.$ $\begin{align} & x<0 \\ & 0\le x\le 1 \\ & x>1 \\ \end{align}$\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,(m{{x}^{2}}+n)\]

=$m{{(0)}^{2}}+n$

=$n$

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,(nx+m)$

=$n{{(0)}^{{}}}+m$

=$m$

Thus, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$exists if m=n

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,(nx+m)$

=$n(1)+m$

=$m+n$

$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,(n{{x}^{3}}+m)$

=$n{{(1)}^{3}}+m$

=$m+n$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(x)$

Thus, $\underset{x\to 1}{\mathop{\lim }}\,f(x)$exists for any internal value of  m and n

Conclusion 

Calculating limits for polynomial and rational functions involves substitution and simplifying expressions. Understanding Left Hand Limits (LHL) and Right Hand Limits (RHL) is crucial for analysing function behaviour at specific points. Limits for trigonometric functions uses geometric proof of the important inequalities relating trigonometric functions. The Sandwich Theorem helps find limits by bounding a function between two known limits. Focus on mastering these methods, as they are fundamental for calculus. Previous year question papers typically have 3-4 questions on limits, emphasizing their importance in exams. Practice these concepts thoroughly to excel in your studies.

Class 11 Maths Chapter 12: Exercises Breakdown

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