# Proving that $f'(0)$ does not exist for this function defined by a series that is both continuous but nowhere differentiable.

So I have the function: $$f(x) = \sum_{n=0}^{\infty} \frac{1}{2^n} h(2^n x)$$ where $h(x)$ is defined as: $$h(x) = |x| \;\ for |x|\leq 1 \;\ and\ h(x+2)=h(x)$$

and what I am trying to do is show that: $f'(0)$ and $f'(1)$ does not exist. For the first derivative at the point $0$ I thought I would let $x = x_m = \frac{1}{2^m}$ and then look at $$\lim \frac{f(x_m) - f(0)}{x_m - 0}$$ but I am not 100% sure how I would evaluate this limit and for $f'(1)$ would I do the same thing just with $f(1)$ instead of $f(0)$ ? Any suggestions would be greatly appreciated.

Cheers.

Note that $x\in [2k-1,2k+1]\Rightarrow h(x)=|x-2k|$. (1)
If $x_{m}=2^{-m}$, by (1) we have $h(2^{n}x_{m})=\left\{\begin{matrix} 0& \text{if} \ n> m\\ 2^{n-m}&\text{if} \ n\leq m \end{matrix}\right.$
Then, $$f(x_{m})=\sum_{n=0}^{m}\frac{h(2^{n-m})}{2^{n}}+ \sum_{n>m}\frac{h(2^{n-m})}{2^{n}}=\sum_{n=0}^{m}2^{-m}=(m+1)2^{-m}$$ So, $$\frac{f(x_{m})-f(0)}{x_{m}-0}= \frac{f(x_{m})}{x_{m}}=(m+1)\rightarrow \infty$$