## University Calculus: Early Transcendentals (3rd Edition)

Prove that for every negative real number $-B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$0\lt|x-3|\lt\delta\Rightarrow f(x)\lt -B$$
*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 $x$ $$0\lt|x-c|\lt \delta\Rightarrow f(x)\lt -B$$ $$\lim_{x\to3}\frac{-2}{(x-3)^2}=-\infty$$ We need to prove here that for every negative real number $-B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$0\lt|x-3|\lt\delta\Rightarrow f(x)\lt -B$$ - Examine the inequality: $$f(x)\lt -B$$ $$\frac{-2}{(x-3)^2}\lt -B$$ $$\frac{2}{(x-3)^2}\gt B$$ $$(x-3)^2\lt\frac{2}{B}$$ $$|x-3|\lt\sqrt{\frac{2}{B}}$$ - So if we set $\delta=\sqrt{\frac{2}{B}}$ here, that would make $0\lt |x-3|\lt\sqrt{\frac{2}{B}}$, then for all $x$, we would have $f(x)\lt -B$. The limit has been proved then.