Find $f'(0)$ if $\frac{f(4h)+f(2h)+f(h)+f\left(\frac{h}{2}\right)+f\left(\frac{h}{4}\right)+f\left(\frac{h}{8}\right)+\cdots}{h}=64$ Assume f is a differentiable function with $f(0)=0$ satisfying the equation 
$$\frac{f(4h)+f(2h)+f(h)+f\left(\frac{h}{2}\right)+f\left(\frac{h}{4}\right)+f\left(\frac{h}{8}\right)+\cdots}{h}=64$$
then find $f'(0)$. This is easy using L'Hopital rule. 
How can one compute this without using L'Hopital?
 A: Replace $h$ with $h/2$ and subtract the resulting equation from given equation to get $f(4h)/h=32$ so that $f(x) =8x,f'(0)=8$. There is no need of differentiability condition.
A: Assume the equation holds for all sufficiently small positive values of $|h|.$
Given $\epsilon>0$ let $\delta>0$ such that $0<|x|<\epsilon\implies |\frac {f(x)}{x}-f'(0)|<\epsilon .$ Then if $|4h|<\delta$ and $2\geq n\in \Bbb Z$ we have $f(h2^n)=h2^nf'(0)(1+R_n)$ where $|R_n|<\epsilon.$ Hence $$\sum_{-\infty}^2f(h2^n)=hf'(0)\sum_{-\infty}^2 2^n+hf'(0)\sum_{-\infty}^22^nR_n=$$ $$ =8hf'(0)+hf'(0)\sum_{-\infty}^22^nR_n.$$ $$\text {Now }\quad |hf'(0)\sum_{-\infty}^22^nR_n|\leq |hf'(0)|\sum_{-\infty}^22^n|R_n|\leq$$ $$\leq |hf'(0)|\sum_{-\infty}^22^n\epsilon=8|hf'(0)|\epsilon.$$ So $|8f'(0)-64|<8|f'(0)| \epsilon,$ where $\epsilon$ is arbitrarily close to $0.$
Some assumptions about the range of $h$ must be made. To conclude $f'(0)=8$ it is not sufficient that the equation holds merely for  $some$  $h\ne 0 $. For example if $f(x)=x^2$ then the LHS is $\frac {64h}{3},$ which is $64$ when $h=3,$ but $f'(0)=0.$ 
