Question

The value of $\displaystyle\lim_{x\to\infty}\left(\frac{\pi}{2} - \tan^{-1} x\right)^{1/x}$ is

• A. 0
• B. 1
• C. -1
• D. e

Answer 1

Answer: (B)

1

Answer 2

Let $y=\displaystyle\lim _{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan ^{-1} x\right)$
Taking log on both sides, we get $\left(\right.$ form $\left.\frac{\infty}{\infty}\right)$
$\log y=\displaystyle\lim _{x \rightarrow \infty} \frac{1}{x} \log \left(\frac{\pi}{2}-\tan ^{-1} x\right)$
$=\displaystyle\lim _{x \rightarrow \infty} \frac{\left(-\frac{1}{1+x^{2}}\right)}{\frac{\pi}{2}-\tan ^{-1} x}\,\,\,\,$ (using L' Hospital's rule)
$=\displaystyle\lim _{x \rightarrow \infty} \frac{\frac{2 x}{\left(1+x^{2}\right)^{2}}}{-\left(\frac{1}{1+x^{2}}\right)} \,\,\,\,$ (using L' Hospital's rule)
$=\displaystyle\lim _{x \rightarrow \infty} \frac{-2 x}{1+x^{2}}=0$
$\Rightarrow y=e^{0}=1$