Proving g(x) is continuous at x = 0

Question

Define $g: \mathbb{R} \to \mathbb{R}$ by $$g(x) =\begin{cases} x^2 & \text{if } x \text{ is rational}, \\\\ x^4 & \text{if } x \text{ is irrational}. \end{cases}$$ Prove that $g$ is continuous at $x = 0$ using $\epsilon - \delta$ definition of continuity.

Is my solution correct?

Solution Attempt:

Let $\epsilon > 0 $

$\lim _{x\to\:a}\:x^{2\:}=\:a^2,\:a=0$

$\left|x^2-a^2|\:=\:\right|x+a\left|\right|x-a\left|\right|$

Suppose $\:\left|x-a\right|<1$

then, $-1<x-a<1 \iff -1+a<x<1+a $

Since $|x+a| \leq |x| + |a|$ , we have $ |x+a| < (2a+1) $

So, $\left|x^2-a^2|\:=\:\right|x+a\left|\right|x-a\left|\right| < |x-a| (2a+1)$

Let $\delta = $ min {$1, \frac{\epsilon}{2a+1}$}

If $|x-a| < \delta$, then

$\left|x^2-a^2|\:=\:\right|x+a\left|\right|x-a\left|\right| < |x-a| (2a+1) < \frac{\epsilon}{2a+1} (2a+1) = \epsilon.$

Answer 1

Let $\epsilon >0$.

$g(0)=0$;

For $|x| <1$:

$|x|^4 \le |x|^2 \le |x|$

Choose $\delta = \min (\epsilon, 1)$.

Then

$|x| \lt \delta$ implies

$|g(x)-g(0)| =|g(x)| \le |x| \lt \epsilon.$

Answer 2

There are two cases. $x=0$ is a rational point, so whenever we look at the region $|c|<\delta$, we can get either very small irrational numbers or very small rationals.

Then we do our estimate $|g(x)-g(c)| = \begin{cases}{ c^2, c \in \mathbb{Q} \\ c^4, c \notin \mathbb{Q}}\end{cases}.$

We must ensure this is less than the given $\epsilon > 0$ in all cases.

You should find a $\delta$ depending on $\epsilon$ for both cases and then take the minimum of both $\delta$.

I.e. you found $\delta = \min(1, \frac{2}{\epsilon+1})$ for rational $c$, now find a similar $\delta$ for irrational $c$.