Question

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

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

Answer 1

The correct option is D $\frac{\pi +2}{4}$

$\underset{x\to 1}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1}{lim}\frac{f^{\prime }\left(x\right)}{1}=f^{\prime }\left(1\right)$

Now $f\left(x\right)=x{\tan }^{-1}x$

$\Rightarrow f^{\prime }\left(x\right)={\tan }^{-1}x+\frac{x}{1+x^2}$

$\therefore f^{\prime }\left(1\right)=\frac{\pi }{4}+\frac{1}{2}=\frac{\pi +2}{4}$