Prove that a function is continuous at $x = x_{0}$ using the $\delta - \epsilon$ definition

Question

Prove that $f(x) = \begin{cases} x^2 & \text{ if } x\in \mathbb{Q} \\ 0 & \text{ if } x\in \mathbb{Q}^{c} \end{cases}$ is continuous at $0$

$\forall \epsilon > 0$, $\exists \delta = ?$ such that

$|x - 0| < \delta$, then

case $1$) if $x$ is rational, then $|f(x) - f(0)| = x^2 < \delta^2$, so take $\delta = \sqrt\epsilon$.

case $2$) if $x$ is irrational, then $|f(x) - f(0)| = 0 < \delta$? What should the $\delta$ be in this case?

Answer 1

If $|x-0| = |x| \leq \delta$ then $$|f(x) - f(0) | = |f(x)| \leq x^2 \leq \delta^2 $$So if $\epsilon > 0$ and $\delta = \sqrt{\epsilon}$ $$|f(x) - f(0) | \leq \epsilon$$Thus $f$ is continous in $0$.