Let $f(x)=x-\left \lfloor{x}\right \rfloor$,
$s(x)=\begin{cases} f(x) & \text{when $f(x) \le0.5$,}\\ 1-f(x) & \text{when $f(x)>0.5$} \end{cases}$,
$s_n(x)=\frac{s(2^{n-1}x)}{2^{n-1}}$,
$b(x)=\sum_{n=1}^{n=\infty}s_n(x)=\sum_{n=1}^{n=N} s_n(x)+\frac{b(2^Nx)}{2^N}$,
$L_N(x)=\sum_{n=1}^{n=N}s_n(x)$ which is linear on interval $[\frac{m}{2^N},\frac{m+1}{2^N}]$ for any integer $m$, and $b(\frac{m}{2^N})=L_N (\frac{m}{2^N})$,
$c$ is the least number in the interval $(a, a+\frac{1}{2^{N-1}}]$ of the form $\frac{m}{2^N}$. Then $d=\frac{m+1}{2^N}$ and $k=\frac{m+1/3}{2^N}$ which is in the interval so $a<c<k<d\le a+\frac{1}{2^{N-1}}$ ,
$L$ is linear function which coincides with $L_N$ on the interval $[c,d]$, By definition $\frac{L(k)-L(a)}{k-a}$=$\frac{L(c)-L(a)}{c-a}$ so If $b(a)\ge L(a)$ then $\frac{b(k)-b(a)}{k-a}$-$\frac{b(c)-b(a)}{c-a}\ge\frac{b(k)-L(k)}{k-a}\ge 1/2$ .
And For case $L(a)\ge b(a)$, $\frac{L(k)-L(a)}{k-a}$=$\frac{L(d)-L(a)}{d-a}$ so
$\frac{b(k)-b(a)}{k-a}$-$\frac{b(d)-b(a)}{d-a}\ge\frac{b(k)-L(k)}{k-a}\ge 1/2$.
The main objective here is to prove that limit of $\frac{b(x)-b(a)}{x-a}$ doesn't exist as $x\to a$. So since We got the inequality for both cases we got $\frac{b(k)-b(a)}{k-a}$-$\frac{b(d)-b(a)}{d-a}\ge\frac{b(k)-L(k)}{k-a}\ge 1/2$ and $\frac{b(k)-b(a)}{k-a}$-$\frac{b(c)-b(a)}{c-a}\ge\frac{b(k)-L(k)}{k-a}\ge 1/2$ from it. I think we can conclude that the limit doesn't exists since $c,k,d$ are in the interval of the neighbourhood of $a$.
Then I checked book's answer which says: Every neighbourhood of $a$ contains an interval of the form $(a,a+1/2^{N-1}]$ and therefore points corresponding to c,k and d. So there can be no limit $g$ such that $|\frac{b(x)-b(a)}{x-a}-g|<1/5$, for all $x$ inside any neighbourhood of $a$.
And my question again arised from the answer of book. Why is $|\frac{b(x)-b(a)}{x-a}-g|$ less than 1/5? Where and how did 1/5 come out of thin air? I think I don't understand the answer of book properly. Why did they choose that the $\epsilon=1/5$ and why not $1/2$? I think $\epsilon$ can't be chosen randomly there must be a way they got it. But which way? So I want to see how to prove that the $b(x)$ is not differentiable more rigorously? I can intutively see that there are infinite amount of sharp turn in self-similar function $b(x)$ which makes it non differentiable but how to prove it is not differentiable using mathematical analysis is somewhat hard for me to do. (If there is some missing information let me know in the comment)