Prove function is differentiable at a point

Question

Prove that the function $$f(x) = \left\{ \begin{array}{ll} x^2, & x \in \mathbb{Q} \\ 0, & x \in \mathbb{Q}^c \\ \end{array} \right.$$

is differentiable at $x = 0.$

I'm not sure how to calculate the limits for this function (the density of $\mathbb{Q}$ in $\mathbb{R}$ is confusing me). Any advice would be appreciated!

Answer 1

We have $$\lim_{x\to 0,x\in \Bbb {Q}}\dfrac{f(x)-f(0)}{x-0}=0$$ and $$\lim_{x\to 0,x\in \mathbb{R}\setminus\mathbb{Q}}\dfrac{f(x)-f(0)}{x-0}=0.$$ Hence, $f$ is differentiable at $x=0$.

Answer 2

We need that

$$\lim\limits_{x \to 0} \frac{f(x)}{x} = 0$$ (having already done some simplification, and noting that the derivative must be zero if it exists, since the function is even).

Now, $$\frac{f(x)}{x} = \left\{\array{x,&x \in \mathbb{Q}\\0,&x \in \mathbb{Q}^c}\right.$$

Clearly, both the functions $g(x) = x$ and $h(x) = 0$ have the limit $0$ at $0$, so by what I guess you might have called the subset limit theorem, so does $\frac{f(x)}{x}$, so the limit is 0. In particular, the limit exists, so $f$ is differentiable at $0$.