Question

Find the value of the given limit, \[\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\operatorname{cotx}-cosx}{{{(\pi -2x)}^{3}}}\].
\[\begin{align}
  & (a)\dfrac{1}{16} \\
 & (b)\dfrac{1}{8} \\
 & (c)\dfrac{1}{4} \\
 & (d)\dfrac{\pi }{2} \\
\end{align}\]

Answer 1

Hint: In the given limit put \[x=\dfrac{\pi }{2}\], if you get the limit as, \[\dfrac{0}{0}\] apply L'Hospital's rule. Differentiate the numerator and differentiate the denominator and then take the limit. Repeat the process till we get the limit as a fraction and not\[\dfrac{0}{0}\].

Complete step-by-step answer:
We have been given an expression with a limit which tends from \[x\]to \[\dfrac{\pi }{2}\].
Now let us equate the given limit to \[l\]. Thus we get,
\[l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\operatorname{cotx}-cosx}{{{(\pi -2x)}^{3}}}\to (1)\]
Now in the above expression let us put, \[x=\dfrac{\pi }{2}\]. Thus we get the value as,
\[\begin{align}
  & l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\operatorname{cotx}-cosx}{{{(\pi -2x)}^{3}}} \\
 & \text{as, }\cot \dfrac{\pi }{2}=0 \\
 &\Rightarrow l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\cot \dfrac{\pi }{2}-cos\dfrac{\pi }{2}}{{{(\pi -2\times \dfrac{\pi }{2})}^{3}}}=\dfrac{0-0}{{{(\pi -\pi )}^{3}}}=\dfrac{0}{0} \\
\end{align}\]
We got the limit as, \[l=\dfrac{0}{0}\]. Thus in cases like these we use the L Hospital’s rule.
Now the L Hospital’s rule states that, if we have an indeterminate form \[\dfrac{0}{0}or\dfrac{\infty }{\infty }\] all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
Now let us differentiate the numerator and denominator of equation (1).
\[l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\operatorname{cotx}-cosx}{{{(\pi -2x)}^{3}}}\]
\[\begin{align}
  & l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\dfrac{d}{dx}(\operatorname{cotx}-cosx)}{\dfrac{d}{dx}{{(\pi -2x)}^{3}}} \\
 &\Rightarrow l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{-\text{cose}{{\text{c}}^{2}}x-(-\sin x)}{3\times {{(\pi -2x)}^{2}}\times (-2)}=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{-\text{cose}{{\text{c}}^{2}}x+\sin x}{(-6)\times {{(\pi -2x)}^{2}}} \\
 &\Rightarrow l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{-\text{cose}{{\text{c}}^{2}}x+\sin x}{(-6)\times {{(\pi -2x)}^{2}}}\to (2) \\
\end{align}\]
Now let us put \[x=\dfrac{\pi }{2}\] in the above expression. Thus we get,
\[\begin{align}
  & l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{-\text{cose}{{\text{c}}^{2}}x+\sin x}{(-6)\times {{(\pi -2x)}^{2}}} \\
 &\Rightarrow l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{-\text{cose}{{\text{c}}^{2}}\dfrac{\pi }{2}+\sin \dfrac{\pi }{2}}{(-6)\times {{(\pi -2\times \dfrac{\pi }{2})}^{2}}}=\dfrac{-1+1}{(-6)\times {{(\pi -\pi )}^{2}}}=\dfrac{0}{0} \\
\end{align}\]
Again we get \[l=\dfrac{0}{0}\]. Hence let us apply L Hospitals rule again and differentiate the numerator and denominator of equation (2).
\[l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\dfrac{d}{dx}(-\text{cose}{{\text{c}}^{2}}x+\sin x)}{\dfrac{d}{dx}(-6)\times {{(\pi -2x)}^{2}}}\]
\[\begin{align}
  & \therefore \dfrac{d}{dx}\text{cose}{{\text{c}}^{2}}x=-2\text{cosec}x(\cot x.\text{cosec}x) \\
 &\Rightarrow l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{-(-2\text{cosec}x(\cot x.\text{cosec}x))+\cos x}{2\times (-6)\times (\pi -2x)\times (-2)}=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{2\text{cosec}x(\cot x.\text{cosec}x)+\cos x}{24(\pi -2x)} \\
\end{align}\]
\[l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{2\text{cosec}x(\cot x.\text{cosec}x)+\cos x}{24(\pi -2x)}\]
\[l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{2\text{cose}{{\text{c}}^{2}}x\cot x+\cos x}{24(\pi -2x)}\to (3)\]
Now let us put \[x=\dfrac{\pi }{2}\] in equation (3). Thus we get the limit as,
\[\begin{align}
  & l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{2\text{cosec}x(\cot x.\text{cosec}x)+\cos x}{24(\pi -2x)} \\
 &\Rightarrow \displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{2\text{cosec}\dfrac{\pi }{2}(\cot \dfrac{\pi }{2}.\text{cosec}\dfrac{\pi }{2})+\cos \dfrac{\pi }{2}}{24(\pi -2\times \dfrac{\pi }{2})}=\dfrac{(-1\times 0)+0}{24(\pi -\pi )}=\dfrac{0}{0} \\
\end{align}\]
Again we get \[l=\dfrac{0}{0}\]. Hence let us apply L Hospitals rule again and differentiate the numerator and denominator of equation (3).
\[\begin{align}
  & l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\dfrac{d}{dx}\left( 2\text{cose}{{\text{c}}^{2}}x\cot x+\cos x \right)}{\dfrac{d}{dx}24(\pi -2x)}\to (3) \\
 &\Rightarrow \therefore \dfrac{d}{dx}2\text{cose}{{\text{c}}^{2}}x\cot x=2\left[ \cot x\dfrac{d}{dx}(cose{{c}^{2}}x)+\text{cose}{{\text{c}}^{2}}x\dfrac{d}{dx}\cot x \right] \\
 &\Rightarrow \therefore \dfrac{d}{dx}2\text{cose}{{\text{c}}^{2}}x\cot x=2\left[ \cot x.(2\operatorname{cosecx}(-cotx.cosecx))+\cos e{{c}^{2}}x.(-\text{cose}{{\text{c}}^{2}}x) \right] \\
 &\Rightarrow \dfrac{d}{dx}2\text{cose}{{\text{c}}^{2}}x\cot x=2\left[ -2{{\cot }^{2}}x\text{cose}{{\text{c}}^{2}}x-\text{cose}{{\text{c}}^{4}}x \right] \\
 &\Rightarrow l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{2\left[ -2{{\cot }^{2}}x\text{cose}{{\text{c}}^{2}}x-\text{cose}{{\text{c}}^{4}}x \right]-\sin x}{24(-2)}\to (4) \\
\end{align}\]
Now let us put \[x=\dfrac{\pi }{2}\] in equation (4 ). Thus we get the limit as,
\[\begin{align}
  & l=\displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{2\left[ -2{{\cot }^{2}}x\text{cose}{{\text{c}}^{2}}x-\text{cose}{{\text{c}}^{4}}x \right]-\sin x}{24(-2)} \\
 &\Rightarrow l=\dfrac{2\left[ -2{{\cot }^{2}}\dfrac{\pi }{2}\text{cose}{{\text{c}}^{2}}\dfrac{\pi }{2}-\text{cose}{{\text{c}}^{4}}\dfrac{\pi }{2} \right]-\sin \dfrac{\pi }{2}}{-48}=\dfrac{2(0-1)-1}{-48}=\dfrac{-3}{-48} \\
 &\Rightarrow l=\dfrac{1}{16} \\
\end{align}\]
Thus we got the required limit as, \[l=\dfrac{1}{16}\].
\[\therefore \displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\operatorname{cotx}-cosx}{{{(\pi -2x)}^{3}}}=\dfrac{1}{16}\]

So, the correct answer is “Option (a)”.

Note: In the case of \[\dfrac{0}{0}\] we typically think of a fraction that has a numerator of zero as being zero. However, we also tend to think of fractions in which the denominator is going to zero, in the limit, as infinity or might not exist at all. Likewise, we tend to think of a fraction in which the numerator and denominator are the same as one. We know \[\dfrac{0}{0}\] can’t exist, hence use L Hospitals rule to solve the same problem.