Question

The value of $\underset{x\to \frac{\pi }{2}}{\lim }{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}$ is

• A. $1$
• B. $0$
• C. $e$
• D. None of these.

Answer 1

The correct option is A

$1$

Explanation of the correct option.

Step $1$: Put the values of limit.

Given :$\underset{x\to \frac{\pi }{2}}{\lim }{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}$

Let $L=\underset{x\to \frac{\pi }{2}}{\lim }{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}$

Taking $\ln$ both side,

$$\begin{gathered} \Rightarrow \ln \left(L\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\ln \left(\sin {\left(x\right)}^{\tan \left(x\right)}\right) \\ \Rightarrow \ln \left(L\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\tan \left(x\right)\ln \left(\sin \left(x\right)\right) \\ \Rightarrow \ln \left(L\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\ln \left(\sin \left(x\right)\right)}{\cot \left(x\right)} \\ \Rightarrow \ln \left(L\right)=\frac{0}{0} \end{gathered}$$

Step $2$: Apply L. Hospital's rule.

Since, it is $\frac{0}{0}$ form, apply L. Hospital's rule,

$\ln \left(L\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{{\displaystyle \frac{1}{\sin \left(x\right)}}.\cos \left(x\right)}{-cose{c}^2\left(x\right)}\right]$

$$\begin{gathered} \Rightarrow \ln \left(L\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{-\cos \left(x\right)\times {\sin }^2\left(x\right)}{\sin \left(x\right)}\right] \\ \Rightarrow \ln \left(L\right)=0 \\ \Rightarrow L=e^0 \\ \Rightarrow L=1 \end{gathered}$$

Therefore, the value of $\underset{x\to \frac{\pi }{2}}{\lim }{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}$ is $1$.

Hence, option $\left(A\right)$ is the correct option, i.e. $1$.