Answer
See explanations.
Work Step by Step
Step 1. To prove the limit $\lim_{x\to3}\frac{-2}{(x-3)^2}=-\infty$, for any large negative number $-B$, we need to find a corresponding value $\delta\gt0$ so that for all x in the interval $0\lt|x-3|\lt\delta$, we get $\frac{-2}{(x-3)^2}\lt -B$
Step 2. The last inequality gives $\frac{2}{(x-3)^2}\gt B$ and $(x-3)^2\lt 2/B$ which leads to $|x-3|\lt\sqrt {2/B}$; thus we can choose $\delta=\sqrt {2/B}$ and when we go back in the steps, for all x in the interval $0\lt|x-3|\lt\delta$, we get $\frac{-2}{(x-3)^2}\lt -B$ which proves the limit statement.