Question

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

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

Answer 1

Answer: (B)

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