Generally speaking, is the following true: $$\lim_{x\to a}f'(x)=\lim_{x\to a}\left(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\right)=\lim_{h\to 0}\left(\lim_{x\to a}\frac{f(x+h)-f(x)}{h}\right)$$

  • $\begingroup$ Note if $f$ is continuous, for each $h\neq 0$, $\lim_{x\to a} \frac{f(x+h) - f(x)}{h} = \frac{f(a+h) - f(a)}{h}$. So the RHS is $f'(a)$. $\endgroup$ – user99914 Nov 21 '17 at 23:53
  • $\begingroup$ Generally speaking, away from the specific circumstances of your equation, limits are not always "commutative", i.e. you cannot always interchange the order of multiple limiting procedures. Usually, when the order of multiple limits cannot be changed, there is some kind of discontinuity or non-smoothness at play. This crops up more frequently in multivariable situations though. en.wikipedia.org/wiki/Iterated_limit $\endgroup$ – jdods Nov 22 '17 at 0:04
  • $\begingroup$ Aside: in latex, instead of ( ... ) you can do \left( ... \right) and it will size the parentheses to fit what's between them. $\endgroup$ – Hurkyl Nov 22 '17 at 0:04
  • $\begingroup$ @jdods so when would I know when it's not "commutative" when dealing with one variable? $\endgroup$ – Tomás Palamás Nov 22 '17 at 1:38
  • $\begingroup$ @Hurkyl thanks! $\endgroup$ – Tomás Palamás Nov 22 '17 at 1:38

Your statement boils down to the continuity of $f'.$

However, the derivative, even if it exists for any point $x\in\mathbb{R},$ does not have to be continuous. A classic counterexample is $f(x)=x^2\sin(1/x)$ for $x\neq 0,$ and $f(0)=0.$

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    $\begingroup$ Technically this is not a counterexample in the sense that $\lim_{x \to 0} f'(x)$ does not even exist. $\endgroup$ – copper.hat Nov 22 '17 at 0:30
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    $\begingroup$ So ... the question is not clear. Does it mean: if all four limits exist, then the two sides are equal? $\endgroup$ – GEdgar Nov 22 '17 at 1:54
  • $\begingroup$ @GEdgar: I guess I would interpret the question as asking if $f$ is such that the two limits exist, are they equal? But then I am a pedant... $\endgroup$ – copper.hat Nov 22 '17 at 2:29
  • $\begingroup$ If f has a derivative at every point and if f' has left and right limits at a point a then these two limits are necessarily equal to f'(a). This is an easy consequence of the intermediate value property of derivatives. $\endgroup$ – Kavi Rama Murthy Nov 22 '17 at 8:17
  • $\begingroup$ @KaviRamaMurthy: Darboux's theorem always surprises me. So, as long as $f'(a)$ exists and the $\lim_{x \to a} f'(x)$ exists then they must be equal, regardless of the lack of assumed continuity of $f'$! $\endgroup$ – copper.hat Nov 22 '17 at 16:25

This is not an answer to the question, but I think worth recording in a less ephemeral way than comments. It is based on an observation by @KaviRamaMurthy.

Suppose $f$ is differentiable in a neighbourhood of $a$ and $\lim_{x \to a} f'(x)$ exists, then this must equal $f'(a)$ as a result of Darboux's theorem.

Note that the only additional assumption to the question is that $f$ is differentiable at $a$, in particular, no continuity is assumed.

To emphasise, if the two limits in the question exist and $f$ is differentiable at $a$ then the limits are equal.


Suppose $\lim_{x \to a} f'(x) = g \neq f'(a)$. Let $\epsilon= { 1\over 2} |g-f'(a)|$ and choose $\delta>0$ such that if $0<|x-a| < \delta$ then $|f'(x)-g| < \epsilon$. Then $f'(B(a,\delta)) \subset \{f'(a)\} \cup B(g,\epsilon)$, note that $f'(a) \notin \overline{B(g,\epsilon)}$.

Pick $x$ such that $0<|x-a|< \delta$. Darboux's theorem states that $(f'(x),f'(a)) \subset f'(B(a,\delta))$ which is a contradiction.

  • $\begingroup$ Why the downvote? $\endgroup$ – copper.hat Nov 22 '17 at 21:27
  • $\begingroup$ Since you mention Darboux's theorem, why don't you write a proof? $\endgroup$ – user99914 Nov 22 '17 at 22:48
  • $\begingroup$ @JohnMa: I added an aside to that effect, but it was really KaviRamaMurthy's suggestion. Did you downvote? $\endgroup$ – copper.hat Nov 23 '17 at 0:38
  • $\begingroup$ I just upvoted your answer (so no, the downvote is not mine. But I can understand the downvote since you did not prove anything). $\endgroup$ – user99914 Nov 23 '17 at 1:37
  • $\begingroup$ @JohnMa: Thanks John. However, this does establish an answer in the affirmative subject to the additional proviso that $f$ is differentiable at $a$. My reluctance was due to the fact that someone else pointed out the use of Darboux's theorem. I wanted to add this because the other answer suggests that continuity is required where in fact we have 'almost continuity' due to the theorem (and added assumption). $\endgroup$ – copper.hat Nov 23 '17 at 3:50

If the limit of derivative exists at a point $a$, then function can be approximated by a line at points really close to left of $a$, now if the limit of derivative exists, then another line with same slope as previous line approximates the points really close to right of $a$. That was about LHS of the equation, now coming to RHS, your first limit means you approach $a$ from any side, and fix an $x$ really close to $a$, with second limit you are talking about derivative at that $x$, which may not exist, but as you get closer to $a$, the derivative is defined at those points close to $a$, because your LHS(limit of derivative exists), meaning derivative exists around a neighborhood of $a$.

As for examples for equality,you can have $y=x$, at non zero $x$, and make this function discontinuous at zero.


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