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

Assume $f'(0)$ exists. Consider $g(x)=(f(x))^3=x$.Then by the chain rule $g'(0)=3(f(0))^2f'(0)$. Since $g'(0)=1$ and $f(0)=0$, we find $1=0\cdot f'(0)$, which is absurd.

For your second question (which you might prefereably have posted separately):
We have
$$\lim_{x\to0^+}e^{-2/x}=0$$
because $-2/x\to-\infty$, but 
$$\lim_{x\to0^-}e^{-2/x}=+\infty$$
because now $-2/x\to+\infty$.
So $\lim_{x\to0}e^{-2/x}$ does not exist.
