$L_1 =(a^nb^n)$ and $L_2 =(a^nb^{2n})$. Is $L_1 \cup L_2$ DCFL?

Question

I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.

Therefore, $L_1 \cup L_2$ is not cfL.

Is that correct?

Answer 1

Hint:

$L_1$ is produced by

S := $\epsilon$ | aSb

$L_2$ is produced by

S := $\epsilon$ | aSbb

(assuming $a^a$ is a typo, meaning $a^n$)