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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 $f'(0) = 0$

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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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    $\begingroup$ You missed to write a step: $f'(0)=\lim_{x\to 0}\displaystyle\frac{f(x)-f(0)}x=\lim_{x\to 0}\frac{f(x)}x\ =\ 0$ by the same argument as your first line. $\endgroup$ – Berci May 9 '14 at 14:41
  • $\begingroup$ Absolutely not. Think e.g. $f(x)=x$. This satisfies $f(0)=0$, while $f'(0)=1$. $\endgroup$ – Berci May 9 '14 at 14:50

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