Error in the derivative proof

Question

Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.

The faulty proof goes as follows:

Let $a\in K$

By the definition of the derivative:

$$h'(a)=\lim_{x\to a}\frac{h(x)-h(a)}{x-a}$$

Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.

$$h'(a)=\lim_{x\to a}\frac{h(x)-h(a)}{x-a}\to(LH)\to \lim_{x\to a}\frac{h'(x)-0}{1-0}=\lim_{x\to a}h'(x)$$

We have proven that $h'(a)=\lim_{x\to a}h'(x)$. By defn, $h'$ is cont.

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There is an error in the proof, and I also need to fix the proof.

I don't see something right away.

My guess is that $x-a\neq 0$, but I am not sure how to prove this.

Answer 1

L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g \to A$.

In your “proof”, $\displaystyle\lim_{x \to a} h'(x)$ need not exist.

(But if it does, then the claim is true: Prove that $f'(a)=\lim_{x\rightarrow a}f'(x)$.)

Answer 2

Let $a=0$, $h(x) = \begin{cases} x^2 \sin {1 \over x} , & x \neq 0 \\ 0, & x = 0\end{cases}$ on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) \over x}$ has no limit as $x \to 0$.