$f''(0)$ exists, what is $\lim_{x\to 0} f(x)/x^2$? Given a function $f:\mathbb{R}\to\mathbb{R}$ twice-derivable at $x=0$ with $f(0)=0$.
Define $g(x):=\frac{f(x)}{x}$ for all $x\neq 0$ and $g(0)=f'(0)$. Clearly, $g$ is continuous since $$ \lim_{x\to 0} g(x) =\lim_{x\to 0}  \frac{f(x)}{x} = \lim_{x\to 0}  \frac{f(x) - f(0)}{x-0} = f'(0).   $$
Solving a problem, I got the limit $\lim_{x\to 0}  \frac{g(x)}{x} =\lim_{x\to 0}\frac{f(x)}{x^2} $, which I can't manage. I suppose that it exists and is related with $f''(0)$. Could anyone show how to compute $\lim_{x\to 0}  \frac{g(x)}{x}$?
Thanks in advance.
EDIT: In a solution-book I read $g'(0)=f''(0) /2$. Can this be  typo? Maybe they forgot some assumptions in the problem.

The example $f(x)=x$ shows that the limit need not exist.
