Question

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

Answer: (A)

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