Question

The value of $\underset{x\to 0}{\lim }\frac{1}{x}\left[{\tan }^{-1}\left(\frac{x+1}{2x+1}\right)-\frac{\pi }{4}\right]$ is

• A. 1
• B. $-\frac{1}{2}$
• C. 2
• D. 0

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }\frac{1}{x}\left[{\tan }^{-1}\left(\frac{x+1}{2x+1}\right)-\frac{\pi }{4}\right]$
$=\underset{x\to 0}{\lim }\left(\frac{1}{x}\right)\left[{\tan }^{-1}\left(\frac{x+1}{2x+1}\right)-{\tan }^{-1}\left(1\right)\right]$
$=\underset{x\to 0}{\lim }\left(\frac{1}{x}\right).{\tan }^{-1}\left(\frac{\frac{x+1}{2x+1}-1}{1+\frac{x+1}{2x+1}}\right)$
$=\underset{x\to 0}{\lim }\frac{1}{x}.{\tan }^{-1}\left(\frac{-x}{3x+2}\right)$
$=-\underset{x\to 0}{\lim }\left[\frac{{\tan }^{-1}\left(\frac{x}{3x+2}\right)}{\frac{x}{3x+2}}\times \frac{1}{3x+2}\right]=-\frac{1}{2}$