Is the Limit of the derivative of a $C^1$ function always the equal to the one-sided derivative? Let $f\colon [0,\newcommand{\eps}{\varepsilon}\eps)\to\newcommand\R{\mathbb R}\R$ and $f\in C^1\bigl((0,\eps)\bigr)$ and the one-sided derivative
$$f'_+(0) = \lim_{h\to +0} \frac{f(h)-f(0)}{h}$$
should exist. Is it always true that it holds
$$ \lim_{s\to+0} f'(s) = f'_+(0) \, ?$$
 A: There are examples where such limit doesn't exist:
$$
f(x)=
\begin{cases}
x^2 \sin(\frac 1 h) & x>0\\
0 & x=0
\end{cases}
$$
We have
$$
f'_+(0)=\lim_{h\to 0^+} \frac{h^2\sin(\frac 1 h)}h=0
$$
and
$$
f'(x)=2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)
$$
so $\lim_{x \to 0} f'(x)$ doesn't exist.
A: $\newcommand{\eps}{\varepsilon}$We additionally assume that $f\in C^2\bigl((0,\eps)\bigr)$ and that 
$f''$ is bounded on $(0,\eps)$.
Proof
The equation we want to show is essentially
$$\begin{align*}
   \lim_{h\to+0} \lim_{s\to+0} \frac{f(s+h) - f(s)}{h} = \lim_{s\to+0} \lim_{h\to+0} \frac{f(s+h) - f(s)}{h}.
\end{align*}$$
We know limits can be interchanged, when both limits exist and one of the
inner limits converges uniformly.
Define for $s>0$ the functions
$$\begin{align*}
   g_h(s) &= \frac{f(s+h) - f(s)}{h} \\
   g(s) &= f'(s) = \lim_{i\to\infty} g_i(s).
\end{align*}$$
Since $f\in C^2\bigl((0,\eps)\bigr)$ we can Taylor's theorem with Lagrange
remainder and get
get
$$
f(s+h) = f(s) + h f'(s) + \frac{h^2}2 f''(\xi_{s,h})
$$
for some $\xi_{s,h}\in(s,s+h)$.
Now consider
$$\begin{align*}
   \sup_{s\in (0,\eps)} |g_h(s) - g(s)| &= \sup_{s\in (0,\eps)} \left|\frac{h}2 f''(\xi_{s,h})\right| \\
     &= h\underbrace{\sup_{s\in (0,\eps)} \left|\frac{1}2 f''(\xi_{s,h})\right|}_{< const.} \overset{h\to 0}{\longrightarrow} 0.
\end{align*}$$
This means that $g_i(s)\to g(s)$ uniformly.
Thus, the limits may be interchanged.

Comments on if this is actually correct and how the premisses may be weakened are very welcome. (As well as additional answers.)
