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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?

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    $\begingroup$ Is it given $f(0)=0$? $\endgroup$
    – jonsno
    Commented May 3, 2018 at 6:34
  • $\begingroup$ It is not given $\endgroup$
    – Anvit
    Commented May 3, 2018 at 6:35
  • $\begingroup$ It is given that it is real valued though $\endgroup$
    – Anvit
    Commented May 3, 2018 at 6:35
  • $\begingroup$ Is it given that $f(x)$ is continuous? $\endgroup$ Commented May 3, 2018 at 6:41
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    $\begingroup$ I'm confused why you assumed that $f$ is analytic $\endgroup$
    – pancini
    Commented May 3, 2018 at 6:42

4 Answers 4

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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

$$\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}h\frac{f(h)}{h^2}=0.$$

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    $\begingroup$ Why is it assumed? It is said in the edit that function is differentiable, so it is already continuous $\endgroup$
    – jonsno
    Commented May 3, 2018 at 7:06
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    $\begingroup$ Yes, that the function is differentiable was said in the edit ! In the original question no such assumption was maid ! $\endgroup$
    – Fred
    Commented May 3, 2018 at 7:43
  • $\begingroup$ I understand but you posted a new answer to address the edited question or not! Anyway answer is fine, so I have upvoted already :) $\endgroup$
    – jonsno
    Commented May 3, 2018 at 7:47
  • $\begingroup$ I proved the following: if $f$ is continuous at $0$ and if $\lim_{x\to0}f(x)/x^2$ exists, then $f$ is differentiable at $0$ and $f'(0)=0$. $\endgroup$
    – Fred
    Commented May 3, 2018 at 8:48
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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.

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  • $\begingroup$ Maybe it is implictly assumed function is continuous? This is from a undergrad admission test. I'm posting question as is $\endgroup$
    – Anvit
    Commented May 3, 2018 at 6:42
  • $\begingroup$ The question is a bad question ! $\endgroup$
    – Fred
    Commented May 3, 2018 at 6:46
  • $\begingroup$ MB, The question says function differentiable. Dunno how i missed it $\endgroup$
    – Anvit
    Commented May 3, 2018 at 6:48
  • $\begingroup$ O.K. Then $f$ is continuous at $0$. Now read my second answer. $\endgroup$
    – Fred
    Commented May 3, 2018 at 6:51
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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

$$\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}h\frac{f(h)}{h^2}=0$$

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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.

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