Proving a function is continuous and differentiable at $0$. The question I have been asked is:

Let $f$ be the function defined by:
$f(x)=\begin{cases}\
        e^{-\frac{1}{x}} & x>0\\
        0 & x \leq 0\\
      \end{cases}$ 
Show that $f$ is continuous at 0 and differentiable at $0$, and also compute the
  derivative $f '(0)$. (Hint: you may use, without proof, the fact that $\frac{e^{-\frac{1}{x}}}{x^n}\to 0$
  as $x → 0$ for any $n ∈ \Bbb{N}$.)

Any help provided would be greatly appreciated. I have a solution but it is very poorly worded, looking for a simpler and more understandable explanation.
Thanks
 A: compute the term $$\frac{f(0+h)-f(0)}{h}$$ and then compute the Limit $$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$ if this Limit exists
A: For the continuity you should verify that:
$$\lim_{x\to0} f(x)=f(x_0)$$
For the derivability that the following limits exist and are equals
$$\lim_{h\to0^+} \frac{f(0+h)-f(0)}{h}=\lim_{h\to0^-} \frac{f(0+h)-f(0)}{h}$$
Note: derivability $\implies$ continuity.
A: $$\lim_{x \to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x \to 0^+}\frac{e^{\frac{1}{x}}}{x}=\lim_{t \to +\infty}te^{-t}=0$$
Where we do the change of variables: $$t=\frac{1}{x} \to +\infty  \text{  as   } x \to 0^+$$
Also $\lim_{x \to 0^-} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0^-}\frac{0}{x}=0$
So both side limits exist at zero and are equal to zero thus $f$ is differentiable at zero  thus continuous at zero and $f'(0)=0$
With  a bit more work you can show that this function is infinitely differentiable at every point.
A: As $x\to0$, $\frac1x\to\infty$, hence $e^{-\frac1x}\to0$. So $f$ is continuous at $0$.
Moreover by L'Hospital's rule $$\lim_{h\to0}\frac{e^{-\frac1h}-0}h=\lim_{h\to0}e^{-\frac1h}\frac1{h^2}=\lim_{x\to\infty}e^{-x}x^2=0$$
So $f'(0)$ exists.
