Answer
See the proof below.
Work Step by Step
We have $$ |x^3+12-12| = |x^3-0| =x^2 |x-0|$$ Since $|x^3+12 -12| $ is a multiple of $|x-0|$, then $|x^3+12 -12| $ is arbitrarily small whenever $ x $ is sufficiently close to $0$. Hence, we get $$\lim_{x\rightarrow 0}x^3+12=12.$$
