derivate of a piecewise function $f(x)$ at$ x=0$. There is a piecewise function $f(x)$
$$f(x)= \begin{cases} 1 ,\ \ \text{if}\ \  x \geq \ 0 \\ 0,\ \  \text{if}\ \  x<0 \end{cases}$$ 
what is the derivative of the $f(x)$ at $x=0$?
Is it $0$? Or since it is not continuous, the derivative does not exists? 
 A: Hint: Prove that if a given function $f$ is differentiable at a point $a$, then $f$ is continuous at $a$.
Sketch: If $f$ is differentiable at $a$ then we may write (Theorem) $f(a+h) = f(a) + hf'(a) + \frac{r(h)}{h}h$ where $lim_{h \to 0 } \frac{r(h)}{h} = 0 $, then $lim_{h\to 0} f(a+h) = f(a)$, that is, $f$ is continuous at $a$. 
Show that your function is not continuous at $0$. Then use the contrapositive $p \Rightarrow q \equiv\  \sim q \Rightarrow\ \sim p$. 
A: Another way you can prove that your function is not differentiable at a=0 is to show that the left hand limit and the right hand limit in the definition of f'(a) are not equal. When you take h approaching 0 from the left, f'(0) is infinite. When you take h approaching 0 from the right, f'(0)=0. Since the left hand and right hand limits are not equal, f'(0) does not exist. 
A: This is the Dirac's delta function.
http://en.wikipedia.org/wiki/Dirac_delta_function
http://mathworld.wolfram.com/DeltaFunction.html
So the answer to your question is that at the origin the derivative is unlimited.
$\lim\limits_{x \rightarrow 0}{\delta(x)}=\infty$
