University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 95

Answer

Prove that for every negative real number $-B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$-\delta\lt x\lt 0\Rightarrow \frac{1}{x}\lt -B$$

Work Step by Step

$$\lim_{x\to0^-}\frac{1}{x}=-\infty$$ According to the formal definition deduced from Exercise 93, we need to prove that for every negative real number $-B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$-\delta\lt x\lt 0\Rightarrow \frac{1}{x}\lt -B$$ - Examine the inequality: $$\frac{1}{x}\lt -B$$ $$x\gt-\frac{1}{B}$$ - So if we set $\delta=\frac{1}{B}$, then we have $-\frac{1}{B}\lt x\lt 0$, meaning that we would for all $x$ have $\frac{1}{x}\lt -B$, satisfying the requirements. The proof has been completed.

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