1

Let $f$ be a differentiable function on an interval containing zero. If $$\lim_{x\to0}f'(x)=l$$ then is it true that $f'(0)=l$.

If $f'$ is a continuous function, then of course it is true. But what if $f'$ is not continuous? Is it still true? I think not, but I am not able to find some example. Any suggestions?

egreg
  • 238,574
Eklavya
  • 2,671
  • 1
    In fact, I think if $f$ is differentiable on $(-a, 0) \cup (0, a)$ for some $a$; $f$ is continuous at $0$; and $f'(x) \to l$ as $x \to 0$ - then $f$ is also differentiable at 0 and $f'(0) = l$. Hint: Mean Value Theorem. – Daniel Schepler Aug 25 '17 at 16:34

2 Answers2

3

If $f$ is differeniable also at zero, then the claim is in fact (perhaps surprisingly) true. The mean value thm tells us that for $x>0$: $$f(x)-f(0)= f'(\xi_x) x$$ where $\xi_x\in (0,x)$. So dividing by $x$ and taking limits $$ f'(0)=\lim_{x\rightarrow 0^+} \frac{f(x)-f(0)}{x} = \lim_{x \rightarrow 0^+} f'(\xi_x) = \ell$$

H. H. Rugh
  • 35,236
  • How did you infer that $\lim_{x \to 0^{+}}f'(\xi_x) = 0$? In the OP the stated limit of $f'(x)$ as $x \to 0$ is an arbitrary $l$. – Chris Aug 25 '17 at 16:44
  • I think it should be an equality on the first line instead of "$\leq$". – nicomezi Aug 25 '17 at 17:03
  • Thanks, corrected. Originally, I thought of writing a proof valid in any dimension (whence the inequality) but anyway a couple of other things should then also be corrected. – H. H. Rugh Aug 25 '17 at 17:42
3

If the function $f$ is continuous at $0$, then the statement is true, because it's a particular case of l’Hôpital’s theorem: you are computing $$ \lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f'(x)}{1} $$ as per l’Hôpital, since the limit on the right hand side is assumed to exist. Continuity of $f$ at $0$ implies that the left hand side is an indeterminate form $0/0$, so the theorem can be applied.

Since you are assuming that $f$ is differentiable, then it is also continuous.

Continuity of $f'$ is not relevant.

egreg
  • 238,574