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Using the method of L’ Hospital, evaluate the limit.

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\log (1-\cot x)}{\cot x}

 

Option: 1

1


Option: 2

0


Option: 3

-1


Option: 4

Cannot be determined


Answers (1)

best_answer

Apply the L' Hospital's Rule, when \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, \text { etc. } [an intermediate form] in the following way

  • Differentiate \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, provided that the limit  \lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}  exists.

  • Otherwise, go on differentiating till the limit with the determinate form is achieved.

The provided limit is

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\log (1-\cot x)}{\cot x}=\frac{0}{0}

Here use the L’ Hospital rule and also refer to the following formula

\begin{aligned} & \frac{d}{d x}(\cot x)=\--cosec ^2 x \\ & \frac{d}{d x}\left(\log _e x\right)=\frac{1}{x} \end{aligned}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}} \frac{\log (1-\cot x)}{\cot x} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{d}{d x} \log (1-\cot x)}{\frac{d}{d x}(\cot x)} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{-\left(-\operatorname{cosec}^2 x\right)}{1-\cot x}}{-\operatorname{cosec}^2 x} \\ & =-\lim _{x \rightarrow \frac{\pi}{2}} \frac{1}{1-\cot x} \quad\left[\frac{\pi}{\triangle} \operatorname{cosec} x \neq 0 \text { as } x \neq \frac{\pi}{2}\right] \\ \end{aligned}

\begin{aligned} & =-\frac{1}{1-\cot \frac{\pi}{2}} \\ & =-1 \end{aligned}

 

Posted by

Pankaj Sanodiya

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