The following function is only continuous at one point. Is it differentiable at that point?

Question

$h(x) = 0, x ∈ \mathbb{R} \setminus \mathbb{Q}$

$h(x) = x^2, x ∈ \mathbb{Q}$

Here is my proof:

$\forall x ∈ \mathbb{Q}, \lim_{x\to 0} x^2 = 0$ $\forall x ∈ \mathbb{R} \setminus \mathbb{Q}, \lim_{x\to 0} 0 = 0$

So the function is continuous at x=0, and h(0)=0

$\forall x ∈ \mathbb{Q}, \lim_{x\to 0} \frac{h(x)-h(0)}{x-0} = \lim_{x\to 0} \frac{x^2}{x} = \lim_{x\to 0} x = 0$

$\forall x ∈ \mathbb{Q}, \lim_{x\to 0} \frac{h(x)-h(0)}{x-0} = \lim_{x\to 0} \frac{0}{x} = 0$

Hence h is differentiable at that point.

Answer 1

It doesn't make sense to write that $(\forall x\in\mathbb{Q}):\lim_{x\to0}\frac{h(x)-h(0)}{x-0}=0$. You can prove that $h$ is differentiable at $0$ by noticing that$$(\forall x\in\mathbb{R}):\left|\frac{h(x)-h(0)}x\right|\leqslant|x|,$$from which it follows that $h'(0)=0$.