## Calculus (3rd Edition)

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 $$\lim_{x\rightarrow 0}x^3+12=12.$$