how to prove $\lim\limits_{x \to 0+} f'(x)$ = $\lim\limits_{x \to 0-} f'(x)$ implies $f'(0)$ exists?

Question

$f:(-1,1) \to R$ is continous and $f'(x)$ exists for all $x \in (-1,1)$ except $x=0$

if $\lim\limits_{x \to 0+} f'(x)$, $\lim\limits_{x \to 0-} f'(x)$ exists and same

how to prove $f'(0)$ exists?

Answer 1

Theorem (Spivak) Suppose $f$ is continuous at $x=a$, that $f'(x)$ exists for all $x$ in a neighborhood of $a$. Suppose moreover that $$\lim_{x\to a}f'(x)$$ exists. Then $f'(a)$ exists and $$f'(a)=\lim_{x\to a}f'(x)$$

Proof By definition, $$f'(a)=\lim_{h\to 0 }\frac{f(a+h)-f(a)}h$$

Consider $h>0$. For $h$ sufficiently small, $f$ will be continuous over $[a,a+h]$, and differentiable over $(a,a+h)$. Thus, by Lagrange, we can find $a<\alpha_h<a+h$ such that $$\frac{f(a+h)-f(a)}h=f'(\alpha_h)$$

As $h\to 0^+$; $\alpha_h\to a$, and since the limit exists, $$f'(a)^+=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}h=\lim_{h\to 0^+}f'(\alpha_h)=\lim_{x\to a}f'(x)$$ The case $h<0$ is analogous. $\blacktriangle$.