University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

Chapter 2 - Questions to Guide Your Review - Page 110: 18

Answer

$\lim_{x\to\infty}f(x)=L$ means that for every number $\epsilon\gt0$, there exists a corresponding number $M$ such that for all $x$, $$x\gt M\Rightarrow |f(x)-L|\lt\epsilon$$ $\lim_{x\to-\infty}f(x)=L$ means that for every number $\epsilon\gt0$, there exists a corresponding number $N$ such that for all $x$, $$x\lt N\Rightarrow |f(x)-L|\lt\epsilon$$

Work Step by Step

1) $\lim_{x\to\infty}f(x)=L$ means that for every number $\epsilon\gt0$, there exists a corresponding number $M$ such that for all $x$, $$x\gt M\Rightarrow |f(x)-L|\lt\epsilon$$ $\lim_{x\to-\infty}f(x)=L$ means that for every number $\epsilon\gt0$, there exists a corresponding number $N$ such that for all $x$, $$x\lt N\Rightarrow |f(x)-L|\lt\epsilon$$ 2) Examples: We examine the function $y=\frac{1}{x}$ As $x\to\pm\infty$, $f(x)$ approaches $0$. So $\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=0$ $\lim_{x\to\infty}f(x)=0$ means that for every number $\epsilon\gt0$, there exists a corresponding number $M$ such that for all $x$, $$x\gt M\Rightarrow |f(x)|\lt\epsilon$$ $$x\gt M\Rightarrow |\frac{1}{x}|\lt\epsilon$$ $$x\gt M\Rightarrow -\epsilon\lt\frac{1}{x}\lt\epsilon$$ $$x\gt M\Rightarrow x\in(-1/\epsilon,0)\cup(1/\epsilon,\infty)$$ So if we take $M=\frac{1}{\epsilon}$, since $x\gt M$, then it is always the case that $|f(x)|\lt\epsilon$. For example, if $\epsilon=2$, then for all $x\gt\frac{1}{2}$, we have $|\frac{1}{x}|\lt2$. Similary, $\lim_{x\to-\infty}f(x)=0$ means that for every number $\epsilon\gt0$, there exists a corresponding number $N$ such that for all $x$, $$x\lt N\Rightarrow |f(x)|\lt\epsilon$$ $$x\lt N\Rightarrow x\in(-1/\epsilon,0)\cup(1/\epsilon,\infty)$$ So if we take $N=\frac{-1}{\epsilon}$, since $x\lt N$, then it is always the case that $|f(x)|\lt\epsilon$. For example, if $\epsilon=2$, then for all $x\lt\frac{-1}{2}$, we have $|\frac{1}{x}|\lt2$.

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