I'm trying to get a deeper understanding of the derivative of a function. I have been reading from the following page:

enter image description here

I have been thinking about why this is an equivalent statement to the original, but I've been having trouble with it.

What I've tried doing so far is by assuming $h \in \mathbb{R}$, and then $\frac{f(x+h) - f(x)}{h}$ is the slope of the secant line from $(x, f(x))$ and $(x+h, f(x+h))$, which could be a decent approximation based on how close $h$ is to zero. But after that I'm having trouble seeing how the little oh gets involved.

Can someone prove to me why the definition of differentiability and the one involving the little-oh notation are equivalent?

  • $\begingroup$ For this exercise, I recommend thinking directly about the definition of "differentiable" and the definition of the little-oh notation; try to show that the last displayed equation both implies the existence of the limit in the differentiability definition and is implied by it (in other words, prove two if-then statements that are converses of each other). There are definitely times to invoke geometric intuition, as you mention; but I believe this is a moment where one should practice understanding and using the explicit rigorous definitions of the terms. $\endgroup$ Commented Mar 14, 2021 at 21:29
  • $\begingroup$ Start with the definition of derivative and prove that the second statement is true. Then assume the second statement is true and show that $f'$ must be the derivative. This should be a straightforward exercise. By the way, this equivalence is what allows us to extend the derivative to multi-dimensions. $\endgroup$
    – John Douma
    Commented Mar 14, 2021 at 22:00
  • $\begingroup$ Thanks for the ideas, I was about to go back to proving it by epsilon delta, but I think @Tortar has a nice proof of it. $\endgroup$ Commented Mar 14, 2021 at 22:19
  • $\begingroup$ Greg, because you mentioned the geometric aspect, do you think you could help me understand the geometric part of the little-oh definition? I've been able to see how it works if I set $f(x) = x^n$ and $g(x) = x^k$ with $k < n$. But not a general geometric view $\endgroup$ Commented Mar 14, 2021 at 22:20
  • $\begingroup$ John, how does it help us generalize to higher dimensions? $\endgroup$ Commented Mar 14, 2021 at 22:20

1 Answer 1


By definition of differentiable function at $x$ you get $$f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} $$

Now if you move the number $f'(x)$ on the other side and inside the limit ( why you can do that?) obtaining $$\lim_{h \to 0}\frac{f(x+h)-f(x)-hf'(x)}{h}= 0 $$

but this is the definition of a function being $o(h)$ so there exists a number $f'(x)$ such that $$o(h) = f(x+h)-f(x)-hf'(x)$$

Rearranging you get the desired expression. (Note that all passages are biconditional)

  • $\begingroup$ I think we can move it inside of the limit because we can rewrite a constant as the limit of $h \to 0$ of itself, then use the sum law for limits. But when they say "if there exists a number $f'(x)$ such that", we are taking that number to be the $f'(x)$ that must exists from the definition of differentiability? $\endgroup$ Commented Mar 14, 2021 at 22:15
  • $\begingroup$ yes that's what I was thinking about, for the question instead note that the statement 'if there exists...' is used in the part with the little oh so you don't have to say that it must exist because of differentiability but because it is assumed to exist otherwise the reasoning would be circular because you would assume differentiability from the start, I edit for clarity $\endgroup$
    – Tortar
    Commented Mar 14, 2021 at 22:19
  • $\begingroup$ When we prove that differentiability implies the little-oh version, we would have to choose that number $f'(x)$, but when we prove that little-oh implies differentiability we would assume it exists. I see that your argument does both at once in a way, but doesn't we have to do two different things depending on which direction we choose? Maybe we could instead prove "if there exists a number $\alpha$ such that" instead so we don't get confused with the $f'(x)$ from the definition of differentiability? $\endgroup$ Commented Mar 14, 2021 at 22:22
  • $\begingroup$ I don't think it matters but I edited to be more clear :) $\endgroup$
    – Tortar
    Commented Mar 14, 2021 at 22:26
  • 1
    $\begingroup$ Ok, no problem. I appreciate your help with this! I just want to ask one thing, when you write $o(h) = ... $ you mean that the thing on the right hand side is a function which is $o(h)$ right? $\endgroup$ Commented Mar 14, 2021 at 22:27

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