$\lim f(x)=0, \lim f'(x)=c$, find $\lim \frac{f(x)}{x}$

Question

PROBLEM:

Let $f:(0,a)\to \mathbb{R} $ be differentiable.

1. $\lim_{x\to 0+} f(x)=0$
2. $\lim_{x\to 0+} f'(x)=c$, for some $c\in\mathbb{R}$

Prove that $\lim_{x\to 0+} \frac{f(x)}{x}=c$

SOURCE:

Real analysis coursebook from Otto Forster, lead up to proving l'hopital's rule

So far I have:

Feels like something very trivial, but I don't know how to write it up or get to the solution. So far I can prove $|\frac{f(x_0)}{x_0}|<c+\epsilon$, but the proof feels very intuition based and non rigorous. Therefore I would rather say I have nothing

Answer 1

Set $g(0)=\lim_{x\to 0+}f(x)$ and $g(x):=f(x)$ for all $a>x\gt 0$.

$g$ is continuous on $[0,a)$. By LMVT on $[0,x]\subset [0,a)$, there is a $c_x\in (0,x)$ such that $$\frac{g(x)-g(0)}{x}=g'(c_x)=f'(c_x)\implies \frac{f(x)}{x}=f'(c_x)$$

The result follows by taking $\lim_{x\to 0+}$ on both sides noting that $c_x\to 0+$ as $x\to 0+$.

Answer 2

You can extend $f$ by continuity to $[0, x]$ and use Lagrange's theorem $$ \dfrac{f(x)-f(0)}{x-0} = f'(\xi(x)), \quad \xi(x) \in (0,x) $$

Since $\xi(x)$ tends to zero when $x\to 0^+$, you have the result.