Let $f$ be a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$.
Suppose that $\lim_{x\rightarrow a}f'(x)$ exists.
Then, prove that $f$ is differentiable at $a$ and $f'(a)=\lim_{x\rightarrow a}f'(x)$.
It seems like an easy example, but a little bit tricky.
I'm not sure which theorems should be used in here.
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Using @David Mitra's advice and @Pete L. Clark's notes
I tried to solve this proof.
I want to know my proof is correct or not.
By MVT, for $h>0$ and $c_h \in (a,a+h)$
$$\frac{f(a+h)-f(a)}{h}=f'(c_h)$$
and $\lim_{h \rightarrow 0^+}c_h=a$.
Then $$\lim_{h \rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h \rightarrow 0^+}f'(c_h)=\lim_{h \rightarrow 0^+}f'(a)$$
But that's enough? I think I should show something more, but don't know what it is.
Best Answer
Some hints:
Using the definition of derivative, you need to show that $$ \lim_{h\rightarrow 0^+} {f(a+h)-f(a)\over h } $$ exists and is equal to $\lim\limits_{x\rightarrow a^+} f'(x)$.
Note that for $h>0$ the Mean Value Theorem provides a point $c_h$ with $a<c_h<a+h$ such that $$ {f(a+h)-f(a)\over h } =f'(c_h). $$
Finally, note that $c_h\rightarrow a^+$ as $h\rightarrow0^+$.