Is f continuous at $ x = \pm 2 $ and everywhere else it isn't continuous?

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be given by $$ f(x) = \begin{cases} x^2 - 2, & x \in \mathbb{I}, \\ -x^2 +2, & x \in \mathbb{Q} \end{cases} $$ Is $f$ continuous at $ x = \pm \sqrt{2} $ and everywhere else it isn't continuous? I think it's easy to prove where it isn't continuous because of the density of the set of reals but how do I prove the continuity in $ x = \pm \sqrt{2}$? I tried it by definition but got stuck.

Answer 1

Note that $f(\sqrt{2}) = 0$, and that $$ x^2-2 = (x-\sqrt{2})(x+\sqrt{2}), $$ so that $$ |-x^2+2 - 0| = |x^2-2 - 0| = |x-\sqrt{2}||x+\sqrt{2}|. $$ Note that when $x$ is close to $\sqrt{2}$, the right factor is close to $\sqrt{2}$, and the left factor is small.

Can you find your delta using this?

Answer 2

You can show this with the $ \epsilon, \delta $ definition by showing that there you can find a $ \delta_1 $ small enough s.t. if $ x\in(\pm2-\delta,\pm2+\delta) $ then for rational x $ |f(x)-f(\pm2)|<\epsilon $ and you can find such a $ \delta_2 $ for irrational x and just calling $ \delta=min\{\delta_1,\delta_2\} $.

Answer 3

$f(x) = 0$ for $x = \pm\sqrt{2}$. Hence we have to show that $f(x) \to 0$ as $x \to\pm\sqrt{2}$. This is equivalent to $\lvert f(x) - 0 \rvert = \lvert x^2 -2 \rvert \to 0$ which is obviuos.