Proving $f'(0)=0$ if $\lim_{x\to0}f(x)/x^2$ exists and is finite Edit: question says the function is differentiable
Given $\lim_{x\to0}f(x)/x^2$ exists and is finite prove $f'(0)=0$
My attempt: $$f(x)=\sum_{k=0}^\infty {f^{n}(0)x^n\over n!}\\
\implies \lim_{x\to0}f(x)/x^2=\frac{f''(0)}2+\lim_{x\to0}\left({f(0)\over x^2}+{f'(0)\over x}\right)$$
How to proceed without l'hopital?
This also assumes function is infnitely differntiable. How to avoid that?
Is the argument that if $f'(0)\ne0$ then $\lim_{x\to0}\left({f'(0)\over x}\right)=\infty$ valid?
 A: If we assume that $f$ is continuous at $0$, then
$$f(0)= \lim_{x \to 0}f(x)= \lim_{x \to 0}x^2 \frac{f(x)}{x^2}=0.$$
Hence
$$\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}h\frac{f(h)}{h^2}=0.$$
A: Consider the function $f : \mathbb R \to \mathbb R$ given by $f(x):=x^2$ if $ x \ne 0$ and $f(0):=1$
Then $\lim_{x\to0}f(x)/x^2=1$, but $f$ is not continuous at $0$, hence $f'(0)$ does not exist.
A: We need to assume continuity here, as per Fred. If $f(0)\neq 0$ then $\lim_{x\to 0}f(x)/x^2$ does not exist. Now
$$\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}h\frac{f(h)}{h^2}=0$$
A: Since it is known that $\lim_{x\to 0}\frac{f(x)}{x^2}=0$ and since $\lim_{x\to 0} x=0$ we get by limit arithmetic that (1) $\lim_{x\to 0} \frac{f(x)}{x} = 0$, Now since it is known that $f’(0)$ exist we also get that $\lim_{x\to 0} \frac{f(x)-f(0)}{x}=f’(0)$ and by subtracting (1) from this limit, We get by limit arithmetic that $\lim_{x\to 0} \frac{-f(0)}{x}=f’(0)$, Now for the limit $\lim_{x\to 0} \frac{-f(0)}{x}$ to exist it must be the case that $f(0)=0$ and so $\lim_{x\to 0} \frac{-f(0)}{x}=\lim_{x\to 0} \frac{0}{x}=0$ and we conclude that $f’(0)=0$ as was to be shown.
