Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.4 - One-Sided Limits - Exercises 2.4 - Page 74: 4

Answer

a. $1$, $1$, $2$ b. Yes. $1$ c. $4$, $4$ d. Yes. $4$

Work Step by Step

Use the figure in the Exercise, given the piecewise function: $f(x)=\begin{cases} 3-x\hspace0.8cm x\lt2 \\2\hspace1.6cmx=2\\ \frac{x}{2}\hspace1.5cm x\gt2 \end{cases}$ a. $\lim_{x\to2^+}f(x)=\frac{2}{2}=1$, $\lim_{x\to2^-}f(x)=3-2=1$, and $f(2)=2$ b. Yes. The limit $\lim_{x\to2}f(x)=1$ because $\lim_{x\to2^+}f(x)=\lim_{x\to2^-}f(x)$ c. $\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}(3-x)=3+1=4$ and $\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}(3-x)=3+1=4$ d. Yes. The limit $\lim_{x\to-1}f(x)=4$ because $\lim_{x\to-1^+}f(x)=\lim_{x\to-1^-}f(x)$
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.