The function $\text{⨍ (x)=}\left\{\begin{array}{ll}{\mathrm{xe}}^{-\left(\frac{1}{\left|\mathrm{x}\right|}+\frac{1}{\mathrm{x}}\right)},& \mathrm{x}\ne 0,\\ 0 , & \mathrm{x}=0\end{array}\right.$ then $\text{⨍ (x)}$ is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\mathrm{xe}}^{-\frac{2}{\mathrm{x}}} ; \mathrm{x}>0\\ \mathrm{x} ; \mathrm{x}<0\\ 0 ; \mathrm{x}=0\end{array}\right. \\ \mathrm{f}\left(0^{-}\right)=\mathrm{f}\left(0^{+}\right)=\mathrm{f}\left(0\right) \\ \therefore \mathrm{f} \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x}=0. \\ {\mathrm{f}}^1\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\mathrm{e}}^{-2/\mathrm{x}}+\mathrm{x}\left({\mathrm{e}}^{-2/\mathrm{x}}.\frac{2}{{\mathrm{x}}^2}\right) ; \mathrm{x}>0\\ 1 ; \mathrm{x}<0 \end{array}\right. \end{gathered}$$

$$\begin{gathered} {\mathrm{f}}^1\left(0^{-}\right)=1, {\mathrm{f}}^1\left(0^{+}\right)=\infty \\ \mathrm{f} \mathrm{is} \mathrm{not} \mathrm{differentiable} \mathrm{at} \mathrm{x}=0 \\ \mathrm{f} \mathrm{is} \mathrm{continuous} \mathrm{for} \mathrm{all} \mathrm{x} \mathrm{but} \mathrm{not} \mathrm{differentiable} \mathrm{at} \mathrm{x}=0 \end{gathered}$$