prove or disprove by counter example If $f$ is differentiable at $x=0$ and $\lim_{x\to0}{f(x)\over x}=3$ then $f(0)=0$ and $f'(0)=3 $. 
so after a few failed counter examples I decided to prove this since it also kinda seems to be true...
this is where I got to:
$f$ is differentiable at $x=0\Rightarrow$ $f'(0)$ exists.
$f'(0)=\lim_{x\to0}{f(x)-f(0)\over x-0}=\lim_{x\to0}{{f(x)\over x}-{f(0)\over x}}$
$f'(0)-3=\lim_{x\to0}({{f(x)\over x}-{f(0)\over x})-\lim_{x\to0}{f(x)\over x}}=\lim_{x\to0}{-{f(0)\over x}}$
$\Rightarrow f'(0)=3-\lim_{x\to0}{{f(0)\over x}}$
now comes the part where i'm not sure about...
$\Rightarrow f'(0)={3\lim_{x\to0}{x\over x}}-\lim_{x\to0}{{f(0)\over x}}={\lim_{x\to0}{3x\over x}}-\lim_{x\to0}{{f(0)\over x}}=\lim_{x\to 0}{{3x-f(0)}\over{x}-0}$
which is by definition the derivative of the function $3x$ at $x=0$ which satisfies both $f'(0)=3$ and $f(0)=0$.... 
is this true? it seems a little stinky since I forced in a specific function....
cheers

In a neighborhood of $x=0$ we have $f(x)=3x+o(x)$, $f$ is differentiable and then continuous at $x=0$ and then $f(x)\to 0$ for $x\to 0$, i.e. $f(0)=0$ and $f'(0)=\lim_{x\to0}{f(x)\over x}=3$.
