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Let the function $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable at $x=0$. Prove that $\lim_{x\rightarrow 0}\frac{f(x^2)-f(0)}{x}=0$.

The result is pretty obvious to me but I am having a difficult time arguing it precise enough for a proof. What I have so far is of course that since $f$ is differentiable; $$f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$$ exists.

Any help would be greatly appreciated.

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5 Answers 5

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HINT:

$$\frac{f(x^2)-f(0)}{x}=\left(\frac{f(x^2)-f(0)}{x^2}\right)x$$

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    $\begingroup$ I wish more Math Overflow answers were like this. There is way too much giving students complete answers to homework problems rather than pointers in the right direction. $\endgroup$
    – John Coleman
    Commented Apr 19, 2018 at 10:54
  • $\begingroup$ @Coleman Thank you. Much appreciated. $\endgroup$
    – Mark Viola
    Commented Apr 19, 2018 at 11:38
  • $\begingroup$ @JohnColeman where would you direct someone to learn how to do proofs like this after failing discrete math twice? $\endgroup$
    – Alex Podworny
    Commented Apr 19, 2018 at 16:32
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    $\begingroup$ @AlexPodworny I would suggest a class in introductory calculus. $\endgroup$
    – Mark Viola
    Commented Apr 19, 2018 at 17:21
  • $\begingroup$ With this approach, after you substitute $y = x^2$ then you have to consider the upper and lower limit separately because of the $x$, right? $\endgroup$
    – Spine Feast
    Commented Apr 20, 2018 at 8:38
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Recall the chain rule, $$ \left(f(g(x))\right)'=f'(g(x))g'(x) $$ here $g(x)=x^2$, and $$ \lim_{x\to 0}\frac{f(g(x))-f(g(0))}{x}=f'(g(0))(g'(0))=f'(0)\cdot 0=0 $$

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Given $\epsilon>0$, there is a $\delta>0$ such that \begin{align*} \left|\dfrac{f(u)-f(0)}{u}-f'(0)\right|<\epsilon/2,~~~~0<|u|<\delta, \end{align*} for all $x$ with $0<|x|<\min\{\sqrt{\delta},1,\epsilon/2(1+|f'(0)|)\}$, then \begin{align*} \left|\dfrac{f(x^{2})-f(0)}{x}\right|&=\left|x\left(\dfrac{f(x^{2})-f(0)}{x^{2}}-f'(0)\right)+xf'(0)\right|\\ &\leq|x|\left|\dfrac{f(x^{2})-f(0)}{x^{2}}-f'(0)\right|+|x||f'(0)|\\ &\leq\epsilon. \end{align*}

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    $\begingroup$ It's great to remind students that they can always go back to the definition of a limit when trying to prove something about a limit. $\endgroup$
    – Eric Lippert
    Commented Apr 19, 2018 at 14:05
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    $\begingroup$ @EricLippert, user284331: The hard part about the definition of a limit isn't merely remembering to use it; it's actually being able to use it. From my personal experience, for students, pulling the right inequalities out of thin air seems to require little short of magic. Here, the last couple lines in this answer very much deserve an explanation. Even though it might seem obvious to you that all that was used here was (I think?) the triangle inequality and Cauchy-Schwarz, neither this fact nor how one would come up with it would be clear to students at this level at all. $\endgroup$
    – user541686
    Commented Apr 20, 2018 at 4:56
  • $\begingroup$ @Mehrdad I completely agree. Recall this earlier answer where the original poster says exactly that -- the steps in the inequality appear to come out of nowhere, even though they are all motivated. math.stackexchange.com/a/2188554/21264 $\endgroup$
    – Eric Lippert
    Commented Apr 20, 2018 at 5:11
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    $\begingroup$ @EricLippert: Haha, wow, I'd completely forgotten about that! I was literally just about to write "didn't realize you were active on Math!" only to realize I already said that there a year ago. :-) Edit: I just read the rest of the comments (I never saw them since the site obviously never notified me) and... yup, exactly. :-) $\endgroup$
    – user541686
    Commented Apr 20, 2018 at 5:17
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If $g(x) = f(x^2)$, then, by the chain rule, $\lim\limits_{x\to 0}\frac{g(x)-g(0)}{x} = \color{blue}{g'(0)} = 2 (\color{red}{0}) f'(0) = \color{red}{0}$.

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To add a little bit of intuition here, consider trying to construct a counterexample. To obtain a limit of 1, instead of 0, we could construct a function where $f(x^2)-f(0) = x$ for all x, and the equation becomes $x/x = 1$ everywhere. The function $f(x) = \sqrt x$ satisfies this for positive x. Allowing negative numbers and ensuring continuity around zero is then satisfied with $f(x) = \begin{cases} \sqrt x & \text{if } x\ge 0 \\ - \sqrt -x & \text{if } x\lt 0 % \end{cases}$

Indeed the limit as x approaches zero is $x / x = 1$. However, it's not a disproof, because in constructing this function, we had to violate the requirements. It has a slope of infinity at $f(0)$ and therefore is not regarded as differentiable at this point, although it is smooth and continuous. Intuitively, any function which tries to violate the stated limit equation must be discontinuous or with infinite slope at zero.

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    $\begingroup$ Is this really an answer to the original question? $\endgroup$
    – Angina Seng
    Commented Apr 19, 2018 at 10:13

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