Show that $f(x)=x^{1/3}$ is not differentiable at $x=0$. 
Show that $$f(x)=x^{1/3}$$ is not differentiable at $x=0$.

$$\text{LHD at} \ x=0$$  $$= \lim_{h\to 0}\frac{f(0-h)-f(h)}{0-h-0}$$$$=\lim_{h\to 0}\frac{-h^{1/3}}{-h}=\lim_{h\to 0}-h^{-2/3}$$
and similarly $$\text{RHD at} \ x=0=\lim_{h\to 0}h^{-2/3}$$
If I directly substitute $h=0$ both will be $0$ or should I take $0$ to the denominator. 
How do I solve this?
and I have another doubt:

How is $$\lim_{x\to 0}e^{-2/x}=0???$$

Thankyou

To show that $f$ is not differentiable at $0$, you need to show that $\frac{f(h) - f(0)}{h}$ does not converge as $h \to 0$ which you did (you get $\pm \infty$ as potential limits).
