Limit of $f'(x)$ - differentiable function

Question

Let $f: [-1,1]$ be a differentiable function such that $\displaystyle\lim_{x \to 0} f'(x) = a$. Show that $f'(0) = a$.

I tried the following:

$a = \displaystyle\lim_{x \to 0} \displaystyle\lim_{y \to x} \dfrac{f(y) - f(x)}{y-x} =$

$= \displaystyle\lim_{x \to 0} \displaystyle\lim_{y \to x} \dfrac{f(y) - f(0)}{y-x} + \displaystyle\lim_{x \to 0} \displaystyle\lim_{y \to x} \dfrac{f(0) - f(x)}{y-x} =$

$= \displaystyle\lim_{y \to x} \displaystyle\lim_{x \to 0} \dfrac{f(y) - f(0)}{y-x} + \displaystyle\lim_{y \to x} \displaystyle\lim_{x \to 0} \dfrac{f(0) - f(x)}{y-x} =$

$= \displaystyle\lim_{y \to 0} \dfrac{f(y) - f(0)}{y} + \displaystyle\lim_{y \to 0} \dfrac{f(0) - f(0)}{y} =$

$= \displaystyle\lim_{y \to 0} \dfrac{f(y) - f(0)}{y} + 0 = f'(0)$

But I don't know if I can exchange the order of limits. Can I do it?

What you wrote in the second line is not valid. The inner limits need not exist. E.g., take $f(x)=x$, $x=1/2$. — David Mitra, Jun 22 '13 at 15:10
If you wish to take advantage of the unnecessary hypothesis that $f'(0)$ exists (I don't think any of the answers to the many duplicates of this question do), suppose $f'(0)\ne a$. Then using the hypothesis that $\lim_{x\rightarrow0} f'(x)=a$, obtain a contradiction to Darboux's Theorem. Of course, this is somewhat silly. Just use MVT for the most elementary proof. — David Mitra, Jun 22 '13 at 15:17
@DavidMitra, very thanks. Surprisingly, 1 line by L'Hopital. And thanks by the counterexample. — Sávio, Jun 22 '13 at 15:18
get the limit of the f'(x) at zero and apply L'Hospital's rule. — Rrjrjtlokrthjji, Jun 22 '13 at 15:18

score 2 · Answer 1 · answered Jun 22 '13 at 15:22

2

An idea:

Directly from definition

$$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}x\stackrel{\text{l'Hospital}}=\lim_{x\to 0}f'(x)\stackrel{\text{given!}}=a$$

answered Jun 22 '13 at 15:22

DonAntonio

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Limit of $f'(x)$ - differentiable function

1 Answers1