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Because $f$ is continuous, by MVT, there is a number d such that $${f(x)-f(c)\over x-c}=f'(d)$$ in (c, x) or (x, c). And if $x\to c$, trivially $d\to c$. So, for $d$ s that have a coresponding x, $f'(d)$ goes to $f'(c)$.

But if $f'(x)$ is not continuous, we can't say $\lim_{x\to c} f'(x)=f'(c)$ generally.

For instance, take $f(x)=x^2\sin (1/x) (x\neq0), 0(x=0)$ and $c=0$, then ${f(x)-f(c)\over x-c}=x\sin(1/x)$ and it goes to $0$ as $x\to 0$. But $f'(x)=2x\sin(1/x)-\cos(1/x)$, so $\lim_{x\to 0} f'(x)$ does not converge and $f'(x)$ is not continuous at $x=0$.

Now I want know if $\lim_{x\to c}f'(x)=\lim_{x\to c} {f(x)-f(c)\over x-c}$ holds when the existence of the value is guaranteed.

I thought that if $f$ is continuous, then $f(x)$ converges at $x=c$ if $f(x)$ converges for some numbers that converge to $c$. But I can't make a formal proof.

The statement to prove is "If $\lim_{x\to c}f'(x)$ exists, the value is $\lim_{x\to c} {f(x)-f(c)\over x-c}$ when f is continuous."

pie
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Zjjorsia
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  • If $\lim_{x \to c}f'(x)=l$ then $\frac {f(x)-f(c)} {x-c}=f'(t_x)$ for some $t_x$ between $c$ and $x$ and $f'(t_x)\to l$ because $t_x \to c$ as $x \to c$. – geetha290krm Mar 27 '24 at 08:01
  • @geetha290krm: this is essentially an answer in a comment. Why not add an answer? – robjohn Mar 27 '24 at 08:21
  • @robjohn Thanks. I have added an answer. – geetha290krm Mar 27 '24 at 08:32
  • See also https://math.stackexchange.com/questions/4517177 and https://math.stackexchange.com/questions/2511227 and https://math.stackexchange.com/questions/3264967 etc. https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3A%24f%27(c)%3D%5Clim_%7Bx%5Cto%20c%7Df%27(x)%24%2C%20OR%20content%3A.&p=1 – Anne Bauval Mar 27 '24 at 09:42
  • @geetha290krm Last but not least: your answer is a duplicate of that one, cleverly avoiding like you the choice of some $t_x$ (as is usually done and as you did in your comment). – Anne Bauval Mar 27 '24 at 10:15

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You have proved the result already but you are just overthinking.

Suppose $\lim_{x\to c}f'(x)=l$. Then, given $\epsilon >0$ there exists $\delta >0$ such that $|f'(x)-l|<\epsilon$ whenever $0 <|x-c|<\delta$.

Let $0 <|x-c|<\delta$. By Mean Value Theorem $\frac {f(x)-f(c)}{x-c}=f'(t)$ for some $t$ between $c$ and $x$. But then $0 <|t-c|<\delta$, so $|f'(t)-l|<\epsilon$. It follows that $|\frac {f(x)-f(c)}{x-c}-l|<\epsilon$ whenever $0 <|x-c|<\delta$ and this finishes the proof.

Note: This also shows that $f$ is differentiable at $c$ and $\lim_{x\to c}f'(x)=f'(c)$ but this has not been asked explicilty.

geetha290krm
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