University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 43


To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-1|\lt\epsilon$$

Work Step by Step

$$\lim_{x\to1}\frac{1}{x}=1$$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-1|\lt\epsilon$$ 1) Solve the inequality $|f(x)-1|\lt\epsilon$ to find an open interval containing $x=1$ on which the inequality holds for all $x\ne1$ $$|f(x)-1|\lt\epsilon$$ $$|\frac{1}{x}-1|\lt\epsilon$$ $$-\epsilon\lt \frac{1}{x}-1\lt\epsilon$$ $$1-\epsilon\lt \frac{1}{x}\lt1+\epsilon$$ $$\frac{1}{1-\epsilon}\gt x\gt\frac{1}{1+\epsilon}$$ So the open interval on which the inequality holds is $(\frac{1}{1+\epsilon},\frac{1}{1-\epsilon})$. 2) Find a value of $\delta\gt0$ that places the centered interval $(1-\delta,1+\delta)$ inside the interval $(\frac{1}{1+\epsilon},\frac{1}{1-\epsilon})$ Take $\delta$ to be the distance from $1$ to the nearer endpoint of $(\frac{1}{1+\epsilon},\frac{1}{1-\epsilon})$. In other words, $\delta=\min(1-\frac{1}{1+\epsilon},\frac{1}{1-\epsilon}-1)$ So if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x-1|\lt\delta$ will automatically place $x$ between $1/(1+\epsilon)$ and $1/(1-\epsilon)$ so that $|f(x)-1|\lt\epsilon$. That means for all $x$, $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-1|\lt\epsilon$$ This completes our proof.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.