Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.3 Exercises - Page 69: 109

Answer

Please see below.

Work Step by Step

We want to prove that $$\lim_{x \to c}[bf(x)]=bL$$ by using $\epsilon - \delta$ definition; that is, we must show that for each $\epsilon >0$, there exists a $\delta >0$ such that $|bf(x)-bL|< \epsilon$ whenever $|x-c|< \delta$. By the assumption, $\lim_{x \to c} f(x) =L$; that is, for any $\epsilon > 0$, there exists a $\delta >0$ such that$$|x-c|< \delta \quad \Rightarrow \quad |f(x)-L|< \epsilon .$$So by choosing $\epsilon = \frac{\epsilon }{|b|}>0$ we conclude that there exists $\delta _1 >0$ such that$$|x-c|< \delta _1 \quad \Rightarrow \quad |f(x)-L|< \frac{\epsilon }{|b|}.$$Now by noting that $|bf(x)-bL|=|b(f(x)-L)|=|b||f(x)-L|$ we conclude that the statement is true.
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.