Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.4 - The Precise Definition of a Limit - 2.4 Exercises - Page 114: 39


$\lim\limits_{x \to 0} f(x)$ does not exist.

Work Step by Step

$f(x) = 0$, if $x$ is rational $f(x) = 1$, if $x$ is irrational Assume $\lim\limits_{x \to 0} f(x) = c$ Let $\epsilon = \frac{1}{3}$ Choose any value for $\delta \gt 0$ Every interval $[0,\delta)$ contains a rational number $a$ and an irrational number $b$ $\vert a - 0 \vert \lt \delta$ and $\vert b - 0 \vert \lt \delta$ Then $\vert f(a) - c \vert \lt \epsilon = \frac{1}{3}$ and $\vert f(b) - c \vert \lt \epsilon = \frac{1}{3}$ Since $f(a) = 0$, then $-\frac{1}{3} \lt c \lt \frac{1}{3}$ Since $f(b) = 1$, then $\frac{2}{3} \lt c \lt \frac{4}{3}$ Clearly this is a contradiction since $c$ can not be in both of these intervals. Therefore, our assumption that $\lim\limits_{x \to 0} f(x) = c$ is false. Thus $\lim\limits_{x \to 0} f(x)$ does not exist.
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.