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

Published by Pearson

# Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 89

#### 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.

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