Question

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

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

Answer 1

Answer: (B)

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