#### Answer

Prove that for every positive real number $B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$0\lt|x+5|\lt\delta\Rightarrow f(x)\gt B$$

#### Work Step by Step

*The formal definition of infinite limits:
$\lim_{x\to c}f(x)=\infty$ if for every positive real number $B$, there exists a corresponding number $\delta\gt0$ such that for all $x$ $$0\lt|x-c|\lt \delta\Rightarrow f(x)\gt B$$
$$\lim_{x\to-5}\frac{1}{(x+5)^2}=\infty$$
We need to prove here that for every positive real number $B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$0\lt|x+5|\lt\delta\Rightarrow f(x)\gt B$$
- Examine the inequality: $$f(x)\gt B$$ $$\frac{1}{(x+5)^2}\gt B$$ $$(x+5)^2\lt \frac{1}{B}$$ $$|x+5|\lt\frac{1}{\sqrt B}$$
- So if we set $\delta=\frac{1}{\sqrt B}$ here, that would make $0\lt |x|\lt\frac{1}{ \sqrt B}$, then for all $x$, we would have $f(x)\gt B$.
The limit has been proved then.