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 $f'(0) = 0$



$\lim\limits_{x\to 0} \frac{f(x)}{x^2}=l$ $\implies$ $\lim\limits_{x\to 0}f(x)=l\times\lim\limits_{x\to 0} x^2$ $\implies$ $\lim\limits_{x\to0}f(x)=0$.

and according to the ability of the differentiation : $\lim\limits_{x\to0}f(x)=f(0)=0$

Hence, $f'(0)=0.$

Notice :

$f'(0)=\lim\limits_{x\to0}\frac{f(x)-f(0)}{x}=\lim\limits_{x\to0}\frac{f(x)}{x}=\lim\limits_{x\to0}\frac{x\times f(x)}{x^2}=\lim\limits_{x\to0}x\times \lim\limits_{x\to0}\frac{f(x)}{x^2}=0\times l=0$

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.