Let $f : \mathbb R\rightarrow\mathbb R$ be a function that is differentiable at zero and such that $f(0)=0$. Show that for each $n\in \mathbb N$ we have that
$$\lim_{x\to0}\frac1x\left(f(x)+f(\frac{x}{2})+...+f(\frac{x}{n})\right)=\left(1+\frac12+...+\frac{1}{n}\right)f'(0)$$
me: i believe that $f'(x)=\frac{f(x)}{x}$ but i still don't understand how to go about this question, any help is appreciated !
$\forall n, \lim_{x\to0}\frac{f(\frac{x}{n})}{x}=\lim_{x\to0}\frac{1}{n}\frac{f(\frac{x}{n})}{\frac{x}{n}}=\frac{1}{n}f^{'}(0)$ by definition of differentiability. Then sum over all $n$, you will get the result.