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I am aware of proofs of this fact, including those given at prove that $f'(a)=\lim_{x\rightarrow a}f'(x)$.. This question is not about how to prove it efficiently using MVT or any other clever method, but whether the proof I've sketched here is correct. This is not a duplicate, because I'm trying to get a review of my delta-epsilon reasoning.

Suppose $f$ is differentiable on an open interval $I$, $c\in I$, and $\lim_{x\to c} f'(x) = L \neq\infty$. Prove that $f'$ is continuous at $c$.

In other words, $f'$ cannot have a removable discontinuity. My intuition is this: if $x$ is close to $c$, then the quantity $\frac{f(x)-f(c)}{x-c}$ is close to $f'(x)$ and also to $f'(c)$, therefore they are close to each other. However, I'm having a hard time formalizing this.

I figure that for any $\epsilon >0$, we can find a $\delta$ so that, whenever $x$ is $\delta$-close to $c$, that puts our difference quotient $\frac{\epsilon}{2}$-close to $f'(c)$. That's true because $f'(c)$ exists.

Additionally, I can find another $\delta$ so that $f'(x)$ is $\frac{\epsilon}{2}$-close to $L$ when $x$ is $\delta$-close to $c$. That's true because the limit of $f'$ exists at $c$.

Choosing the smaller of the two $\delta$'s, the triangle inequality gives us that $|f'(c)-L|<\epsilon$, and so they're equal? Does that work?

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  • $\begingroup$ Mean Value Theorem. $\endgroup$ – David C. Ullrich Oct 4 '15 at 22:38
  • $\begingroup$ fyi this is stated as a theorem and proved in Spivak's Calculus $\endgroup$ – K_P Oct 4 '15 at 22:48
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    $\begingroup$ Well your proof is not correct. You need to show that $f'(c) =L$. You have shown that the different quotient is near $f'(c) $ and the derivative $f'(x) $ is near $L$. But you have not shown how we can get $f'(x) $ close to the difference quotient. $\endgroup$ – Paramanand Singh Jun 30 '17 at 11:26
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    $\begingroup$ Note that the typical proof by MVT establishes that if $x$ is near $c$ then difference quotient is near $f'(c)$ and further that this quotient is equal to the derivative $f'(z) $ for some $z$ where $z$ is even nearer to $c$ than $x$. Therefore $f'(z) $ is near to $L$ and combining this with the information in last sentence we see that the difference quotient is also near $L$ (apart from being already near to $f'(c) $) . This is not possible unless $L=f'(c) $. $\endgroup$ – Paramanand Singh Jul 1 '17 at 7:31
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    $\begingroup$ I hope you can formalize the argument of my last comment using $\epsilon, \delta$ without any problem. $\endgroup$ – Paramanand Singh Jul 1 '17 at 7:35
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On OP's request I convert my comments into an answer.


While the argument put forth by OP in his question is almost correct, it needs to be substantiated by some amount of rigor. And precisely that is the reason it can't be considered as a correct proof.

Note that since $$\lim_{x\to c} \frac{f(x) - f(c)} {x-c} =f'(c) $$ it is correct to say that the difference quotient $(f(x) - f(c)) /(x-c) $ is near $f'(c) $ when $x$ is near $c$. And by interchanging the roles of $x$ and $c$ OP reasons that the quotient is also near $f'(x) $. This is a fundamental flaw but subtle to detect. The amount of closeness between $x$ and $c$ which is needed to ensure the closeness of the difference quotient to the value $f'(c) $ is not necessarily the same as that needed for making the difference quotient close to $f'(x)$. This is difficult to justify rigorously and ultimately it requires some sort of continuity of derivative at $c$ and this is what we actually need to prove.

The simplest approach is to use the mean value theorem and one of the answers has already given this approach. I will also use the same approach bu use the $\epsilon, \delta$ gymnastics. We are supposed to prove that $$\lim_{x\to c} \frac{f(x) - f(c)} {x-c} =f'(c) =L=\lim_{x\to c} f'(x)\tag{1} $$ where first and last equalities are given in the question and the middle one is to be established on the basis of those.

One of the techniques commonly used in analysis to establish the equality of two real numbers is to show that their difference is less than any arbitrarily given positive real number. So let $\epsilon>0$ be arbitrary and we will show that $|L-f'(c) |<\epsilon$. From the first equality in $(1)$ we know that there is a number $\delta_{1}>0$ such that $$\left|\frac{f(x) - f(c)} {x-c} - f'(c) \right|<\frac{\epsilon} {2}\tag{2}$$ whenever $0<|x-c|<\delta_{1}$. And similarly there is a number $\delta_{2}>0$ such that $$|f'(x) - L|<\frac{\epsilon} {2} \tag{3}$$ whenever $0<|x-c|<\delta_{2}$.

Let $\delta=\min(\delta_{1},\delta_{2})$ and let $x$ be any value satisfying $0<|x-c|<\delta$. Clearly the inequality $(2)$ holds for this value of $x$. By mean value theorem we see that there is a number $y $ between $x$ and $c$ such that $$\frac{f(x) - f(c)} {x-c} =f'(y) $$ and then the inequality $(2)$ becomes $|f'(y) - f'(c) |<\epsilon/2$. Since $y$ lies between $x$ and $c$ we have $$0<|y-c|<|x-c|<\delta\leq \delta_{2}$$ From inequality $(3)$ it now follows that $|f'(y) - L|<\epsilon /2$. We can now see via triangle inequality that $$|L-f'(c)|\leq |L-f'(y) |+|f'(y) - f'(c) |<\frac{\epsilon} {2}+\frac{\epsilon}{2}=\epsilon$$ and our proof is complete.

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By mean value theorem, there is a $|c_x-c|<|x-c|$ such that $$\frac{f(x)-f(c)}{x-c}=f'(c_x),$$ and thus $$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}f'(c_x)= L.$$

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What you want to prove is that $f'(c)=L$. But $$ f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h} $$ and, since $f$ is continuous at $c$ (because it's differentiable at $c$), we can use l'Hôpital's theorem: $$ f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}f'(x) $$ provided the last limit exists.

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