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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.

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.

K Split X
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2 Answers2

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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 $f'(a)=\lim_{x\rightarrow a}f'(x)$.)

Hans Lundmark
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  • By need not exist, do you mean infinity? – K Split X Jan 08 '19 at 20:44
  • No, I mean “not exist”. Like for the standard example $h(x) = x^2 \sin(x)$ for $x \neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x \to 0$, so that there is no limit (neither finite nor infinite). – Hans Lundmark Jan 08 '19 at 20:47
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    Oops, that should have been $h(x)=x^2 \sin(1/x)$, of course. – Hans Lundmark Jan 08 '19 at 20:53
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    +1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity. – Paramanand Singh Jan 09 '19 at 19:13
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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 $x \to 0$.

copper.hat
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