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