Question

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

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

Answer 1

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

Given,
$f\left(x\right)=x{\tan }^{-1}\frac{1}{x}$ for $x\ne 0$
and $f\left(0\right)=0$
$\because -\frac{\pi }{2}\le {\tan }^{-1}\frac{1}{x}\le \frac{\pi }{2}$
$\therefore -\frac{\pi }{2}x\le x{\tan }^{-1}\frac{1}{x}\le \frac{\pi }{2}x$
Here, $\underset{x\to 0}{lim}x{\tan }^{-1}\frac{1}{x}=\underset{x\to 0}{lim}\frac{{\tan }^{-1}\frac{1}{x}}{\frac{1}{x}}=0$
And $f\left(0\right)=0$
$\therefore$ $f\left(x\right)$ is continuous at $x=0$
But $\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0}{lim}{\tan }^{-1}\frac{1}{x}$ does not exist
$\therefore$ $f\left(x\right)$ is not differential at $x=0$