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 - Review Exercises - Page 91: 10

Answer

$$\lim_{x \to 5} 9=9$$

Work Step by Step

$$\lim_{x \to 5} 9=9$$Now, we want to prove this limit by using $\varepsilon - \delta$ definition; that is, we must show that for each $\varepsilon >0$, there exists a $\delta >0$ such that $|9-9|< \varepsilon$ whenever $|x-5|< \delta$. Now, the inequality $0=|9-9| < \varepsilon$ always holds for any positive real number $\varepsilon$. So, by choosing any positive real number $\delta >0$ (for example, $1$, $\epsilon$ , ...), we conclude that$$|x-5|< \delta \quad \Rightarrow \quad |9-9|=0 < \varepsilon . $$Done.
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