Show that the following function is infinitely continuously differentiable $$
f(x) = \begin{cases}
  e^\frac{1}{x} &\text{für } x<0 \\    
  0             &\text{für } x \geq 0
\end{cases} 
$$
Problem: I know that to first I have to show that the function is differentiable at $f(x) = 0$ and $f(x) = e^{\frac{1}{x}}$.
Difference quotient for $f(x)=0$ is for me clear, but for $f(x) = e^{\frac{1}{x}}$, I was only able to get to this point:
$$
\lim_{h\to 0} \frac{f(h+x)-f(x)}{h} 
= \lim_{h\to 0} \frac{e^{\frac{1}{h+x}} - e^{\frac{1}{x}}}{h} ={} ?
$$
And also I don't understand how to show in general that a function is infinitely continuously differentiable.  I will be very happy for help.

well we can use the chain rule to differentiate $e^{1/x}$: we get $\frac{d}{dx}[e^{1/x}] = - \left(\frac{1}{x^2}\right) * e^{1/x}$, and from here it is obvious that we can differentiate the function as many times as we want (infinitely differentiable), as each new derivative will contain the term $e^{1/x}$ times some expression with (1/x) raised to some power.
