Question

Show that the function $f\left(x\right)$ defined as $f\left(x\right)=x\cos \frac{1}{x},x\ne 0$, $=0,x=0$ is continuous at $x=0$ but not differentiable at $x=0$.

Answer 1

We have,

$f\left(x\right)=x\cos \frac{1}{x},x\ne 0$

$=0x=0$

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}x\cos \left(\frac{1}{x}\right)=0=f\left(0\right)$

Hence,

$f\left(x\right)$ is continuous at $x=0$.

$f^{\prime }\left(0^{+}\right)=\underset{x\to 0}{lim}f\left(x\right)-f\left(0\right)$

$=\underset{x\to 0}{lim}\frac{h\cos \left(\frac{1}{h}\right)=0}{hh}=\underset{x\to 0}{lim}\cos \frac{1}{h}$=A not fixed number

$f^{\prime }\left(0^{+}\right)$ Does not exist.

Hence,

$f\left(x\right)$ is continuous at $x=0$ but not differentiable at $x=0$.