Continuous $f:[0,1]\to\mathbb{R}$ is differentiable on $(0,1]$ and $\lim_{x\to 0^+} f'(x)$ is finite. Prove that f has right hand derivative at x=0

Question

Suppose $f:[0,1]\to\mathbb{R}$ is differentiable on $(0,1]$ and $\lim_{x\to 0^+} f'(x)$ exists and it is finite. Prove that $f$ has a right hand derivative at $x=0$.

What I know is that there exists some c such that $$\lim_{x\to 0^+}\left(\lim_{x\to y} \frac{f(x)-f(y)}{x-y}\right)=c$$ and $$\lim_{x\to a} f(x)=f(a)$$

I really don't see how I should start from here, any hints welcome.

Answer 1

For $x \in (0,1]$ we have, by the mean value theorem:

$ \frac{f(x)-f(0)}{x-0}=f'(c(x))$ for some $c(x) \in (0,x)$.

If $L= \lim_{t\to0^+} f'(t$), then we get $ \frac{f(x)-f(0)}{x-0}=f'(c(x)) \to L$ as $x \to 0^+$

Answer 2

$$f'(0)=\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}f'(\xi_{x})=\lim_{\xi_{x}\to 0^+}f'(\xi_{x}).$$