# 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.

## marked as duplicate by Hans Lundmark, Jonas Dahlbæk, Trevor Gunn, dantopa, Yujie ZhaJun 29 '17 at 15:58

• You are on the right track. You need to apply the Mean Value Theorem if you have seen that. – Geoff Robinson Jan 13 '13 at 18:59
• Here is a related post. – David Mitra Jan 13 '13 at 19:16

Apply l'Hopital's rule to your quotient $$\frac{f(x)-f(0)}{x}.$$
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 $x<0$
Yes, as Julien's answer says, you can apply L'Hopital. Alternatively, you can use the Mean value theorem directly, which is really how L'Hopital's theorem is proved. For any $h \neq 0,$ we have $\frac{f(x+h)- f(x)}{h} = f^{\prime}(\alpha_{h})$ for some $\alpha_{h} \in (x+h,x)$ for $h <0$, or in $(x,x+h)$ if $h >0.$ Now as $h \to 0,$ we must have $\alpha_{h} \to 0.$ But you are told that $f^{\prime}(\alpha_{h}) \to 1$ as $\alpha_{h} \to 0,$ so you are done
If $\displaystyle\lim_{x\to0}f'(x)=1$ and $f'(0) \not= 1$ then $f'$ has a simple discontinuity at $x=0$ but since $f$ is continuous $f'$ cannot have simple discontinuities then $f'(0)=1$