Question

The value of $\underset{x\to \infty }{\lim }{\left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(x\right)\right]}^{\frac{1}{x}}$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $e$

Answer 1

The correct option is B

$1$

Explanation of the correct option.

Step $1$: Put the values of limit.

Given : $\underset{x\to \infty }{\lim }{\left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(x\right)\right]}^{\frac{1}{x}}$

Let $L=\underset{x\to \infty }{\lim }{\left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(x\right)\right]}^{\frac{1}{x}}$,

taking $\ln$ both sides,

$\ln \left(L\right)=\underset{x\to \infty }{\lim }\ln \left({\left(\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(x\right)\right)}^{\frac{1}{x}}\right)$

$$\begin{gathered} \Rightarrow \ln \left(L\right)=\underset{x\to \infty }{\lim }\frac{1}{x}\ln \left(\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(x\right)\right) \\ \Rightarrow \ln \left(L\right)=\underset{x\to \infty }{\lim }\frac{1}{x}\ln \left({\cot }^{-1}\left(x\right)\right) \\ \Rightarrow \ln \left(L\right)=\frac{\infty }{\infty } \end{gathered}$$

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

Since, it is $\frac{\infty }{\infty }$ form, applying L.Hosptal's rule,

$\ln \left(L\right)=\underset{x\to \infty }{\lim }\frac{{\displaystyle \frac{1}{{\cot }^{-1}\left(x\right)}}\left(-{\displaystyle \frac{1}{1+x^2}}\right)}{1}$

$$\begin{gathered} \Rightarrow \ln \left(L\right)=-\underset{x\to \infty }{\lim }\frac{{\left(1+x^2\right)}^{-1}}{{\cot }^{-1}\left(x\right)} \\ \Rightarrow \ln \left(L\right)=\frac{0}{0} \end{gathered}$$

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

$\ln \left(L\right)=\underset{x\to \infty }{\lim }\frac{{\displaystyle \frac{-2x}{{(1+x^2)}^2}}}{{\displaystyle \frac{-1}{1+x^2}}}$

$$\begin{gathered} \Rightarrow \ln \left(L\right)=-2\underset{x\to \infty }{\lim }\left(\frac{x}{1+x^2}\right) \\ \Rightarrow \ln \left(L\right)=\frac{\infty }{\infty } \end{gathered}$$

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

$\ln \left(L\right)=-2\underset{x\to \infty }{\lim }\frac{1}{2x}$

$$\begin{gathered} \Rightarrow \ln \left(L\right)=-2\left(0\right) \\ \Rightarrow \ln \left(L\right)=0 \\ \Rightarrow L=e^0 \\ \Rightarrow L=1 \end{gathered}$$

Therefore the value of $\underset{x\to \infty }{\lim }{\left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(x\right)\right]}^{\frac{1}{x}}$ is $1$.

Hence, option (B) is the correct option, i.e. $1$.