Question

Examine the origin for continuity and derivability in case of the function $f$ defined by $f\left(x\right)=x{\tan }^{-1}\left(1/x\right)$, $x\ne 0$ and $f\left(0\right)=0$.

• A. continuous but not differentiable at $x=0$
• B. continuous and differentiable at $x=0$
• C. not continuous and not differentiable at $x=0$
• D. none of these

Answer 1

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

$f\left(0^{+}\right)=\underset{h\to 0}{lim}\left(hta{n}^{-1}\right(\frac{1}{h}\left)\right)=0\times \frac{\pi }{2}=0$
$f\left(0^{-}\right)=\underset{h\to 0}{lim}(-hta{n}^{-1}(\frac{1}{-h}\left)\right)=0\times \frac{-\pi }{2}=0$
Hence the function is continuous at $x=0.$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{hta{n}^{-1}\left(\frac{1}{h}\right)}{h}=\frac{\pi }{2}$
$f\left(0^{+}\right)=\underset{h\to 0}{lim}(-hta{n}^{-1}(\frac{1}{-h}\left)\right)=\frac{-\pi }{2}$
Hence the function is not differentiable at $x=0.$