Question

The function $f\left(x\right)=x{\tan }^{-1}\frac{1}{x}$ for $x\ne 0,f\left(0\right)=0$ is :

• A. continuous at $x=0$ but not differentiable at $x=0$
• B. differentiable at $x=0$
• C. neither continuous at $x=0$ nor differentiable at $x=0$
• D. not continuous at $x=0$

Answer 1

The correct option is A continuous at $x=0$ but not differentiable at $x=0$

${\tan }^{-1}x\in (\frac{-\pi }{2},\frac{\pi }{2}),\forall x\in R$
$\Rightarrow \underset{x\to 0}{lim}x{\tan }^{-1}\frac{1}{x}=0$
$\therefore f\left(x\right)$ is continuous at $x=0$

$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h-0}$
$\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}{\tan }^{-1}\frac{1}{h}=\frac{\pi }{2}$

$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{f(-h)-f\left(0\right)}{-h-0}$
$\Rightarrow f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}{\tan }^{-1}\left(\frac{1}{-h}\right)=\frac{-\pi }{2}$

$f^{\prime }\left(0^{-}\right)\ne f^{\prime }\left(0^{+}\right)$
So, $f\left(x\right)$ is not differentiable at $x=0$