real analysis - If the condition of differentiability holds for the rationals then the function is differentiable?

$f:\mathbb{R}\rightarrow\mathbb{R}$ continuous, $a\in\mathbb{R}$. Suppose that there exists $L\in\mathbb{R}$ such that for every $\varepsilon>0$ there exists $r(\varepsilon)>0$ such that $|\frac{f(x)-f(a)}{x-a}-L|<\varepsilon$ for every $x\in\mathbb{Q}$ and $|x-a|

Answer

Fix a point $x\neq a$ such that $0 < |x - a| < r(\epsilon/2)$. Then, at the point x, the function $$g(y) = \frac{f(y) - f(a)}{y - a}$$ is continuous. Now, pick a point $p \in Q$ near $x$ such that $0< |p-a| < r(\epsilon/2)$ and $|g(p) - g(x)| < \epsilon/2$. This is possible by continuity. Now, use your condition to conclude via the triangle inequality, that
$$ \left|\frac{f(x) - f(a)}{x - a} - L\right| = |g(x) - L| \le |g(x) - g(p)| + |g(p) - L| < \epsilon/2 + \epsilon/2 = \epsilon.$$

It follows by definition that $f$ is differentiable at $a$ with $f'(a) = L$.