How to show $f'(0)$ exist and is equal to $1$? [duplicate]

Assume that $f$ be continuous on $\mathbb{R}$, $f'(x)$ exists for all $x\neq 0$, and $\lim_{x\rightarrow 0} f'(x)=1$. We need to show $f'(0)$ exist and is equal to $1$.

$f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$, $\lim_{x\rightarrow 0}f'(x)=1\Rightarrow\lim_{x\rightarrow 0}\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=1$...am I going in the right direction? Please help.

Solution 1:

Apply l'Hopital's rule to your quotient $$ \frac{f(x)-f(0)}{x}. $$

Solution 2:

Let $x>0$. By the Mean Value Theorem in $[0,x]$, $$\exists \xi_x\in (0,x):f'(\xi_x)=\frac{f(x)-f(0)}{x}$$ Let $\epsilon>0$. We have that $$\exists \delta>0: \ \forall y\in (0,\delta)\ \left|f'(y)-1\right|<\epsilon\implies \left|f'(\xi_x)-1\right|<\epsilon\implies \left|\frac{f(x)-f(0)}{x}-1\right|<\epsilon $$ for $0<x<\delta$. Thus $f'_+(0)=1$. Similarly deal with the case $x<0$