(c) If $f$ is differentiable on an interval containing zero and if $\lim_{x→0} f'(x) = L$, then $f'(0) = L$.

1. How to presage proof by contradiction?

Proof by contradiction. True. Assume that $L \neq f'(0)$ and choose e > 0 so that $|f'(0) − L| > e$.
From the hypothesis that $\lim_{x→0} f'(x) = L$,
we know there exists a $δ > 0$ such that $0 < |x| < δ$ implies that $|f'(x) − L| < e$.
Now our choice of $e$ guarantees that there exists a point $f'(0) < a < \color{red}L$ but $|a - L| > e$.

2. How? We chose $e$ guarantee this?

However, by Darboux’s Theorem, there exists a point x such that $|x| < d$ such that $f(x) = a$.

3. How? $\color{red}L$ not a derivative of $f$? Ergo what sanctions use of Darboux's Theorem here?

This suggests that $a - L< e$, a contradiction.

4. What suggests this?

  • $\begingroup$ Your method is good, but with minor mistakes. I have put a proof along the same approach. $\endgroup$ – Paramanand Singh Feb 27 '14 at 8:15

Let's try another way and salvage the proof which is suggested by OP. Suppose $\lim_{x \to 0}f'(x) = L$ and $f'(0) \neq L$. Also assume that $f'(0) < L$. The case $f'(0) > L$ can be handled similarly. let's take an $\epsilon$ with $0 < \epsilon < (L - f'(0))/2$. Now we have a $\delta > 0$ such that $$L - \epsilon < f'(x) < L + \epsilon\tag{1}$$ for all $x$ satisfying $0 < |x| < \delta$. Clearly by the definition of $\epsilon$ we can choose a number $A$ such that $f'(0) < A < L - \epsilon$. If $0 < |x| < \delta$ then we have $f'(0) < A < L - \epsilon < f'(x)$ and by Darboux theorem we have a value $c$ between $0$ and $x$ such that $f'(c) = A$.

Now the contradiction is obvious. We have $0 < |c| < \delta$ and $f'(c) = A < L - \epsilon$ and this contradicts $(1)$.

Note: What we have effectively proved is that any function possessing intermediate value property can not have jump discontinuities.


According to Mean Value Theorem $$ \frac{f(h)-f(0)}{h}=f'(\vartheta h), $$ for some $\vartheta\in (0,1)$. But as $h\to 0$, the right-hand side of the above tends to $\lim_{h\to 0}f'(h)$. Therefore, $f$ the limit as $h\to 0$, of the left-hand side also exists, which means that $f$ is differentiable at $0$, and it's derivative is equal to $\lim_{h\to 0}f'(h)$.


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