If f(x)=≤ft […] ., then f(x) is

We have,

$f(x)=\left\{\begin{array}{cc} x e^{-\left(-\frac{1}{x}+\frac{1}{x}\right)}=x & , x<0 \\ x e^{-2 / x} & , x>0 \\ 0 & , x=0 \end{array}\right.$

Clearly, $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0} x=0$

$\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0} x e^{-2 / x}=0 \times 0=0$ and, f(0) = 0

∴  $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)$

So, f(x) is continuous at x = 0.

Also,

⇒ (LDH at x = 0) = $\left(\frac{d}{d x}(x)\right)_{\text {at } x=0}=1$

⇒ (RHD at x = 0) = $\left\{\frac{d}{d x}\left(x e^{-2 / x}\right)\right\}_{\text {at } x=0}$

⇒ (RHD at x = 0) = $\left\{2^{-2 / x}+\frac{2 e^{-2 / x}}{x}\right\}_{\text {at } x=0}$, which does not exist

Thus, f(x) is everywhere continuous but not differentiable at x = 0.