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  • <img alt="\lim _{x \rightarrow \pi} \frac{\cos ^{-1}(\cos x)-\pi}{\sin ^{-1}(\sin x)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow%20%5Cpi%7D%20%5Cfrac%7B%5C

\lim _{x \rightarrow \pi} \frac{\cos ^{-1}(\cos x)-\pi}{\sin ^{-1}(\sin x)}  is equal to


 

Option: 1

-1


Option: 2

1


Option: 3

\pi


Option: 4

 None of these


Answers (1)

best_answer

Given limit is \lim _{x \rightarrow \pi} \frac{\cos ^{-1}(\cos x)-\pi}{\sin ^{-1}(\sin x)}.
Since, \cos ^{-1}(\cos x)=\left\{\begin{array}{cc}x, & \text { if } 0<x<\pi \\ 2 \pi-x, & \text { if } \pi<x<2 \pi\end{array}\right.
Hence, we have to evaluate the \mathrm{LHL} and RHL, separately.
\begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow \pi^{-}} \frac{\cos ^{-1}(\cos x)-\pi}{\sin ^{-1}(\sin x)}=\frac{x-\pi}{\pi-x}=-1 \\ & \begin{aligned} \mathrm{RHL} & =\lim _{x \rightarrow \pi^{+}} \frac{\cos ^{-1}(\cos x)-\pi}{\sin ^{-1}(\sin x)} \\ & =\frac{(2 \pi-x)-\pi}{\pi-x}=1 \end{aligned} \\ & \text { LHL } \neq \mathrm{RHL} \end{aligned}

So, the given limit does not exists.

 

Posted by

Suraj Bhandari

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