Spivak: Determine if a function $F(x)=\int_0^x f$ is differentiable at $0$. $f$ is a function given by a particular graph. 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?
 A: Pretty much. There are a couple of little errors that can be corrected easily enough. They are just typos, stylistic errors, and small oversights, so I do think your intuition is perfectly fine.
First, there's what I think is a typo:

$$\frac{F(1/2^n)}{2^n}=\frac{1}{2}$$

should be

$$\frac{F(1/2^n)}{1/2^n}=\frac{1}{2}.$$

Second, you should introduce $h$ before you write:

There exists an $n\in\mathbb{N}$ such that $\frac{1}{2^n}<h$.

Before this, it was a dummy variable. What you should write instead is,

Given any $h > 0$, there exists an $n\in\mathbb{N}$ such that $\frac{1}{2^n}<h$.

Third, the fact that $F(x) / x = 1/2$, for our constructed $x \in (0, h)$, does not quite imply that $\lim_{h \to 0^+} F(h) / h$ does not exist! Instead, it proves that if it exists, it must be $1/2$. This is sufficient, however, because $\lim_{h \to 0^-} F(h) / h$ is $0$, as you showed. If the two-sided limit existed, the one-sided limits would have to both exist and agree, which would make $\lim_{h \to 0^+} F(h) / h = 0$. Your argument proves this is impossible.
Finally, I thought I'd point out why the sum is $\frac{1}{2^n}$. The series you're looking at here is telescoping! It is geometric too, but let's do it with telescoping series. When we expand any partial sum, most of the terms cancel:
\begin{align*}
\sum_{i=n}^N \frac{1}{2^i} - \frac{1}{2^{i+1}} &= \frac{1}{2^n} - \frac{1}{2^{n+1}} + \frac{1}{2^{n+1}} - \frac{1}{2^{n+2}} + \frac{1}{2^{n+2}} - \frac{1}{2^{n+3}} + \ldots + \frac{1}{2^N} - \frac{1}{2^{N+1}} \\
&= \frac{1}{2^n} - \frac{1}{2^{N+1}}
\end{align*}
As $N \to \infty$, this approaches $\frac{1}{2^n}$.
A: Take the sequence  $\,y_{n}$ of points such that $f(y_{n})=1$. It turns out that
$y_{n}=3/2^{n+1}$. Now consider $\int_{0}^{y_{n}}f(t)dt.$ It is easy to check that
$\int_{0}^{y_{1}}f(t)dt=\frac{1}{8}(1+\frac{1}{2}+\ldots)=\frac{1}{4}.\,\,$  Likewise
$\int_{0}^{y_{2}}f(t)dt=\frac{1}{16}(1+\frac{1}{2}+\ldots)=\frac{1}{8}$ and $\,\,\int_{0}^{y_{n}}f(t)dt=1/2^{n+1}$.
If we consider the ratio $\int_{0}^{y_{n}}f(t)dt/y_{n}$ we see
that is $1/3$ and hence the limit is $1/3$ which implies that
lim$(F(y_{n})/y_{n})$ is NOT zero as it should be if $F$ was differentiable at $x=0$.
Therefore the function $F$ is not differentiable at $x=0$!
