#### Answer

Prove that for every negative real number $-B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow f(x)\lt-B$$

#### Work Step by Step

*The formal definition of infinite limits:
$\lim_{x\to c}f(x)=-\infty$ if for every negative real number $-B$, there exists a corresponding number $\delta\gt0$ such that for all $x$ $$0\lt|x-c|\lt \delta\Rightarrow f(x)\lt-B$$
$$\lim_{x\to0}\frac{-1}{x^2}=-\infty$$
We need to prove here that for every negative real number $-B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow f(x)\lt-B$$
- Examine the inequality: $$f(x)\lt-B$$ $$-\frac{1}{x^2}\lt-B$$ $$\frac{1}{x^2}\gt B$$ $$x^2\lt\frac{1}{B}$$ $$-\frac{1}{\sqrt B}\lt x\lt\frac{1}{\sqrt B}$$
- So if we set $\delta=\frac{1}{\sqrt B}$, that would make $0\lt |x|\lt\frac{1}{\sqrt B}$, then for all $x$, we would have $f(x)\lt-B$.
The limit has been proved then.