Question

The function $f(x) = x \, \tan^{-1} \frac{1}{x} $ for $x \neq 0 , f(0) = 0$ 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 $x \neq 0$
and $f(0)=0$
$\because \,-\frac{\pi}{2} \leq \tan ^{-1} \frac{1}{x} \leq \frac{\pi}{2}$
$\therefore \, -\frac{\pi}{2} x \leq x \tan ^{-1} \frac{1}{x} \leq \frac{\pi}{2} x$
Here, $\displaystyle\lim _{x \rightarrow 0} x \tan ^{-1} \frac{1}{x}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\tan ^{-1} \frac{1}{x}}{\frac{1}{x}}$
$=0$
$\left[\because \lim _{t \rightarrow 0} \frac{x}{t}\right]$
And $f(0)=0$ (given)
$\therefore f(x)$ is continuous at $x=0$
But, $\displaystyle\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$
$=\displaystyle\lim _{x \rightarrow 0} \tan ^{-1} \frac{1}{x}$ does not exist.
$\therefore f(x)$ is not differential at $x=0$.