Is the function differentiable Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that for $x_0 \in \mathbb{R}$
$$ \lim_{\mathbb{Q} \ni h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$$
exists. Is this function differentiable at $x_0$?
 A: Let 
$$ L=\lim_{\mathbb Q\ni h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$$
Let $(h_n)_{n\in\mathbb N}$ be a sequence with $h_n\to 0$. 
We need to show that 
$$\lim_{n\to\infty}\frac{f(x_0+h_n)-f(x_0)}{h_n}=L.$$
Consider $$\begin{align}g\colon\mathbb R\setminus\{0\}\to&\mathbb R\\ h\quad\mapsto &\frac{f(x_0+h)-f(x_0)}{h}.\end{align}$$
Then $g$ is continuous and hence $|g(h_n')-g(h_n)|<\frac1n$ holds for all $h_n'\ne 0$ sufficiently close to $h_n$. Since the rationals are dense in $\mathbb R$, we thus can find $h_n'$ such that $h_n'\in\mathbb Q\setminus\{0\}$ and $|h_n'-h_n|<\frac1n$ and $|g(h_n')-g(h_n)|<\frac1n$. Then $\mathbb Q\ni h_n'\to 0$, hence $g(h_n')\to L$ and $g(h_n)\to L$ as was to be shown.
A: Suppose that the limit $ \displaystyle L := \lim_{\substack{h \to 0; \\ h \in \mathbb{Q}}} \frac{f(x_{0} + h) - f(x_{0})}{h} $ exists.

Claim: $ f $ is differentiable at $ x_{0} $.

Proof: Let $ (h_{n})_{n \in \mathbb{N}} $ be a sequence in $ \mathbb{R} \setminus \{ 0 \} $ that converges to $ 0 $. It suffices to prove that the limit
$$
\lim_{n \to \infty} \frac{f(x_{0} + h_{n}) - f(x_{0})}{h_{n}}
$$
exists.

For each $ n \in \mathbb{N} $, use the continuity of $ f $ and the denseness of $ \mathbb{Q} $ to choose a $ q_{n} \in \mathbb{Q} \setminus \{ 0 \} $ arbitrarily close to $ h_{n} $ so that


*

*$ \forall n \in \mathbb{N}: \quad |f(x_{0} + h_{n}) - f(x_{0} + q_{n})| < \left| \dfrac{h_{n}}{n} \right| $ and

*$ \displaystyle \lim_{n \to \infty} \frac{q_{n}}{h_{n}} = 1 $, which automatically yields $ \displaystyle \lim_{n \to \infty} q_{n} = 0 $.

Next, observe that
\begin{align}
\forall n \in \mathbb{N}: \quad
   \frac{f(x_{0} + h_{n}) - f(x_{0})}{h_{n}}
&= \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} +
   \frac{f(x_{0} + q_{n}) - f(x_{0})}{h_{n}} \\
&= \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} +
   \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \cdot \frac{q_{n}}{h_{n}}.
\end{align}

As
$$
\forall n \in \mathbb{N}: \quad
  \left| \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} \right|
< \left| \frac{1}{h_{n}} \cdot \frac{h_{n}}{n} \right|
= \frac{1}{n} \stackrel{n \to \infty}{\longrightarrow} 0,
$$
the Squeeze Theorem yields
$$
\lim_{n \to \infty} \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} = 0.
$$

Therefore,
\begin{align}
   f'(x_{0})
= &\lim_{n \to \infty} \frac{f(x_{0} + h_{n}) - f(x_{0})}{h_{n}} \\
= &\lim_{n \to \infty} \left[ \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} +
   \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \cdot \frac{q_{n}}{h_{n}} \right] \\
= &\lim_{n \to \infty} \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} +
   \left[ \lim_{n \to \infty} \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \right]
   \left( \lim_{n \to \infty} \frac{q_{n}}{h_{n}} \right) \\
= &0 + \left[ \lim_{n \to \infty} \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \right]
  \cdot 1 \quad (\text{By the previous paragraphs.}) \\
= &\lim_{n \to \infty} \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \\
= &L. \quad (\text{By the initial hypothesis.}) \quad \spadesuit
\end{align}
A: This is Hagen von Eitzen's proof, but without sequences.
We may assume $x_0=f(x_0)=0$. The function $g(h):={f(h)\over h}$ $\ (h\ne0)$ is continuous, and after subtracting a linear function from $f$ our basic assumption is   $$\lim_{\Bbb Q\ni h'\to 0}g(h')=0\ .$$
Given an $\epsilon>0$ there is a $\delta>0$ such that
$$\left|g(h')\right|<\epsilon\qquad\bigl(|h'|<\delta,\ h'\in\Bbb Q)\ .$$
Assume now that $0<|h|<\delta$. Since $g$ is continuous at $h$ there is a $h'\in\Bbb Q$ with $0<|h'|<\delta$ and $|g(h')-g(h)|<\epsilon$. It follows that $|g(h)|<2\epsilon$. This proves $\lim_{h\to0} g(h)=0$, since $\epsilon>0$ was arbitrary.
