Answer
See the proof below.
Work Step by Step
We have $$ |3x^2-9-(-9)| =|3x^2-9+9| = |3x^2-0| =3x|x-0|$$
Because $|3x^2-9 -(9)| $ is a multiple of $|x-0|$, then $|3x^2-9 -(9)| $ is arbitrarily small whenever $ x $ is sufficiently close to $0$. Hence, we get $$\lim_{x\rightarrow 0}3x^2-9=-9.$$