Prove that if $\lim_{x\rightarrow a}f'(x)$ exists and is finite for all $a \in \mathbb R$,then $f'$ is continous

Question

Let $f:\mathbb R\rightarrow\mathbb R$ be a differentiable function and let $f':\mathbb R\rightarrow\mathbb R$ be the derivative of f.Prove that if $\lim_{x\rightarrow a}f'(x)$ exists and is finite for all $a \in \mathbb R$,then $f'$ is continous.

May I ask for a formal $\epsilon-\delta$ proof?

Thanks in advance!

Answer 1

A proof without $\epsilon-\delta$ (might not be acceptable for you).

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

By the Mean Value Theorem there exists $c_x\in(a,x)$ if $x>a$ or $c_x\in(x,a)$ if $x<a$ such that

$f(x)-f(a)=f'(c_x)(x-a)$

Moreover $x\rightarrow a$ implies $c_x\rightarrow a$ (by squeeze theorem). So we have:

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

So $f'$ is continuous at $a$.

Answer 2

By Darboux Theorem $f'$ has the intermediate value theorem.

Now, assume by contradiction that $f$ is not continuous at some point $a$. Then, since the limit exists, we have $$l=\lim_{x \to a} f'(x) \neq f'(a)$$

Let $$\epsilon =\frac{|f'(a)-l|}{3}$$

Then, there exists some $\delta$ so that $0<|x-a|<\delta$ implies that $$\left| f'(x)-L \right| <\epsilon$$

Now, pick one such $x$ and apply the Intermediate Value Property between $x$ and $a$ for some $M$ close enough to $M$.

Answer 3

Fix $\epsilon > 0$ and $a\in \Bbb R$. Let $L = \lim_{x\to a} f'(x)$. There exists a $\delta_1 > 0$ such that for all $x$, $0 < \lvert x - a\rvert < \delta_1$ implies $\lvert f'(x) - L\rvert < \epsilon$. As $f$ is differentiable at $a$, there exists a $\delta_2 > 0$ such that for all $x$, $0 < \lvert x - a\rvert < \delta_2$ implies $\lvert f(x) - f(a) - f'(a)(x - a)\rvert < \epsilon\lvert x - a\rvert$. By the mean value theorem, for every $x$ there is an $a_x$ in between $x$ and $a$ such that $f(x) - f(a) = f'(a_x)(x - a)$. Thus $\lvert f'(a_x) - f'(a)\rvert < \epsilon$ whenever $0 < \lvert x - a\rvert < \delta_2$. Let $\delta = \min\{\delta_1, \delta_2\}$. If $0 < \lvert x - a\rvert < \delta$, then $0 < \lvert a_x - a\rvert < \lvert x - a\rvert < \delta_1$, and so $\lvert f'(a_x) - L\rvert < \epsilon$; by the triangle inequality,

$$\lvert f'(x) - f'(a)\rvert \le \lvert f'(x) - L\rvert + \lvert L - f'(a_x)\rvert + \lvert f'(a_x) - f'(a)\rvert < \epsilon + \epsilon + \epsilon = 3\epsilon.$$

Since $\epsilon$ and $a$ were arbitrary, the result follows.