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Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0\in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0\neq x\in(a,b)$. Is the following statement true:

If $\lim\limits_{x\to x_0}f'(x)$ exists, then $f'(x_0)$ exists and $\lim\limits_{x\to x_0}f'(x)=f'(x_0)$.

Thanks!

snulty
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boaz
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  • The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous. – Masacroso Jun 20 '16 at 11:36
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    No, this is weaker than saying $f'$ is continuous. – GEdgar Jun 20 '16 at 21:07

4 Answers4

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A qualified "yes": If $f$ is continuous at $x_{0}$, and if $\lim\limits_{x \to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.

Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)

The claim follows from the Mean Value Theorem: If $\delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < \delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that $$ \frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t). $$ Since $|t - x_{0}| < |h|$, taking the limit as $h \to 0$ forces $t - x_{0} \to 0$, as well, so $$ f'(x_{0}) = \lim_{h \to 0} \frac{f(x_{0} + h) - f(x_{0})}{h} = \lim_{t \to x_{0}} f'(t). $$

(Continuity of $f$ was needed to invoke the Mean Value Theorem.)

Andrew D. Hwang
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    Thanks Andrew, a great answer. – boaz Jun 20 '16 at 12:04
  • You're welcome. :) (And, fixed a couple of typos.) – Andrew D. Hwang Jun 20 '16 at 20:03
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    This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2\sin(x), x \ne 0, f(0)=0$ differentiable everywhere? – GEdgar Jun 20 '16 at 21:10
  • Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like. – dfnu Dec 18 '18 at 09:42
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An easier proof, for the chosen answer, is by using the L'Hopital's rule. We know that $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. Now if f is continuous at $a$ the we have a $\frac{0}{0}$ situation, and we can apply the L'Hopital's rule to see that if the limit of $f(x)$ when $x\mapsto a$ exists then it is equal to $f'(a)$.

mosen
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No, it is not true. For a simple counterexample take the indicator function $\chi_{[0,1)}$ in $x\in(-1,1)$ and take $x_0$ to be 0.

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Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $\lim_{x\to x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.

Here's a counter-example: Consider the function:

$$f(x) = \begin{cases} 0, & \mbox{if } \mbox{$x \le 0$} \\ x, & \mbox{if } \mbox{$x>0$} \end{cases}$$

Obviously $f$ is defined on $\mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.

Stefan4024
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  • In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...? – Andrew D. Hwang Jun 20 '16 at 11:56
  • @AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other. – Stefan4024 Jun 20 '16 at 11:59
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    Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”. – egreg Jun 20 '16 at 21:25
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    This is an incorrect answer to the question. $\lim_{x\to 0} f'(x)$ does not exist in your example. – zhw. Jun 20 '16 at 21:39