prove $\lim_{x\to 0} f(x) = 0$ using epsilon delta

Question

prove using Epsilon Delta that $\lim_{x\to 0} f(x) = 0$, where $f(x) = \left\{ \begin{array}{l l} \;\;\; \sqrt6 \;x & \quad \text{if } {x \in \mathbb{Q}}\\ -\sqrt6 \;x & \quad \text{if } {x \not\in \mathbb{Q}} \end{array} \right.$

I know that in order to prove it, I must use: $$\delta = \frac{\epsilon}{\sqrt6}$$

Would the proper proof method be that I must show that the $\delta$ value is the same regardless of if $x$ is rational or irrational?

i.E. Do I show the following?

When $x \in \mathbb{Q}:$ $$0 < \; \mid x \mid \; < \delta \implies \mid \sqrt6 \;x \mid \; < \sqrt6 \; \delta = \epsilon$$

When $x \not\in \mathbb{Q}:$ $$0 < \; \mid x \mid \; < \delta \implies \mid -\sqrt6 \;x \mid \; < \sqrt6 \; \delta = \epsilon$$

Of course, I am not saying the above is the complete proof, but is that the correct strategy in proving it? I haven't done epsilon-delta proofs on a piece-wise function before, which is why I am asking.

Answer 1

Yes, regardless of whether $x\in\Bbb Q$ or not, $|f(x)|=|\pm x\sqrt6|=|x|\sqrt6$.