Question

If $f(x)=x{\tan }^{-1}x$, then $\underset{x\to 1}{\lim }\frac{f(x)-f(1)}{x-1}$ equals to

• A. $\frac{\pi +3}{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi +1}{4}$
• D. $\frac{\pi +2}{4}$

Answer 1

Answer: (D)

Given, $f(x)=x{\tan }^{-1}x$ then
Then, $\underset{x\to 1}{\lim }\frac{f(x)-f(1)}{x-1}(\frac{0}{0}$ form $)$
$=\underset{x\to 1}{\lim }\frac{f^{{}^{\prime }}(x)-0}{1}$(using L'Hospital rule)
$=\underset{x\to 1}{\lim }\left(\frac{x}{1+x^2}+{\tan }^{-1}x\right)$
$=\frac{1}{1+1^2}+{\tan }^{-1}=\frac{1}{2}+\frac{\pi }{4}$
$=\frac{2+\pi }{4}$ or $\frac{\pi +2}{4}$