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

The statement is true. $f$ is differentiable at 0 implies that
$$
f(x) = f'(0)x + f(0) + r(x),
$$
where $r(x)$ is a remainder term such that $\lim_{x\rightarrow 0}\frac{r(x)}{x} = 0.$
Thus
$$
\lim_{x\rightarrow 0}\frac{f(x)}{x} = \lim_{x\rightarrow 0}\left(f'(0) + \frac{f(0)}{x}+ \frac{r(x)}{x}\right).  
$$
On LHS we get 3. On the RHS we have
$$
f'(0) + \lim_{x\rightarrow 0}\frac{f(0)}{x}.
$$
If $f(0) \neq 0$, then this term would be infinite, giving a contradiction.
