Question

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

• A. 0
• B. 1
• C. -1
• D. e

Answer 1

Answer: (B)

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