TITLE: Functions, limits and derivatives
QUESTION [0 upvotes]: Let $f(x)$ be a real valued differentiable function on the real line $\mathbb{R}$ such that when
$\lim\limits_{x\to0} \frac{f(x)}{x^2}$ exists, and is finite. Prove that $f'(0) = 0$

REPLY [1 votes]: Well,
$\lim\limits_{x\to 0} \frac{f(x)}{x^2}=l$ $\implies$ $\lim\limits_{x\to 0}f(x)=l\times\lim\limits_{x\to 0} x^2$ $\implies$ $\lim\limits_{x\to0}f(x)=0$.
and according to the ability of the differentiation : $\lim\limits_{x\to0}f(x)=f(0)=0$
Hence, $f'(0)=0.$
Notice :
$f'(0)=\lim\limits_{x\to0}\frac{f(x)-f(0)}{x}=\lim\limits_{x\to0}\frac{f(x)}{x}=\lim\limits_{x\to0}\frac{x\times f(x)}{x^2}=\lim\limits_{x\to0}x\times \lim\limits_{x\to0}\frac{f(x)}{x^2}=0\times l=0$
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