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Let $f:\ \mathbb{R} \rightarrow \mathbb{R}:\ x \mapsto f(x)$ be a function that is derivable on $\mathbb{R}_0$. Suppose that we know that $\lim_{x \rightarrow 0}f'(x)$ exists and is finite. Can we conclude that $f$ is derivable in $0$?

What I'm trying to prove is that the following limit exists: $$\lim_{x \rightarrow 0}\frac{f(x)-f(0)}{x}$$ I have feeling that we're going to require uniform continuity somewhere. I started out by writing out the limit in full:

$$\lim_{a \rightarrow 0}f'(a) = \lim_{a \rightarrow 0}\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}$$

Now, can I just say the following or are there preconditions? $$\lim_{a \rightarrow 0}\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} = \lim_{x \rightarrow 0}\frac{f(x)-f(0)}{x}$$

Syd Kerckhove
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1 Answers1

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Switching two limits is a rather delicate issue. As a rule, you need some form of uniformity of one limit. However your question has an easier answer: apply the MVT, so that there exists $\theta$ between $a$ and $x$ such that $$ f(x)=f(a)+f'(\theta)(x-a). $$ Hence $\theta \to a$ as $x \to $, and you conclude that $f'(a)=\lim_{x \to a} f'(x)$.

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