The following is a problem from Ch. 14 "The Fundamental Theorem of Calculus", from Spivak's Calculus
Let $F(x)=\int_0^x f$.
Consider the function depicted in the picture below
At which points $x$ is $F'(x)=f(x)$?
The solution manual says
All $x\neq 0$. $F$ is not differentiable at $0$ because $F(x)=0$ for $x\leq 0$, but there are $x>0$ arbitrarily close to 0 with $\frac{F(x)}{x}=\frac{1}{2}$
I'd like to understand this solution better. Here is my attempt at filling in the intermediate steps in the proof.
$f$ is continuous everywhere except at $x=0$. Thus, we can apply the first fundamental theorem of calculus to conclude that
$$F'(x)=f(x)$$
when $x\neq 0$.
$f$ also happens to be integrable everywhere.
What happens at $x=0$ is the question.
If $F$ is differentiable at $0$ then the following limits exist and are equal to each other
$$\lim\limits_{h\to 0^-} \frac{F(h)}{h}=\lim\limits_{h\to 0^+} \frac{F(h)}{h}$$
Now
$$\lim\limits_{h\to 0^-} \frac{F(h)}{h}=0$$
Do we also have $\lim\limits_{h\to 0^+} \frac{F(h)}{h}=0$?
There exists an $n\in\mathbb{N}$ such that $\frac{1}{2^n}<h$.
The sum of areas of triangles formed by $f$ up to $\frac{1}{2^n}$ is
$$F(1/2^n)=\frac{1 \cdot \sum\limits_{i=n}^\infty \left ( \frac{1}{2^i}-\frac{1}{2^{i+1}}\right )}{2}$$
I'm not sure what the infinite sum is, but I'm guessing it is $\frac{1}{2^n}$
Thus
$$\frac{F(1/2^n)}{2^n}=\frac{1}{2}$$
That is, there is always $x$ such that $0<x<h$ and
$$\frac{F(x)}{x}=\frac{1}{2}$$
Therefore, $\lim\limits_{h\to 0^+} \frac{F(h)}{h}$ doesn't exist.
Hence $F$ is not differentiable at $0$.
Is this correct?