Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 1 - Ingredients of Change: Functions and Limits - 1.3 Activities - Page 31: 13

Answer

$$ \lim _{x \rightarrow 5 } \frac{2 x-10}{x-5}=2 $$

Work Step by Step

Given $$ \lim _{x \rightarrow 5} \frac{2 x-10}{x-5} $$ We use a table: \begin{array}{|c|c|c|c|c|}\hline x \to5^-& {4.9} & {4.99} & {4.999} & {4.9999} \\ \hline \frac{2 x-10 }{ x-5} & {2} & {2} & {2} & {2} \\ \hline\end{array} This means that $$ \lim _{x \rightarrow 5^-} \frac{2 x-10}{x-5}=2 $$and \begin{array}{|c|c|c|c|c|}\hline x\to5^+ & {5.1} & {5.01} & {5.001} & {5.0001} \\ \hline \frac{2 x-10 }{ x-5} & {2} & {2} & {2} & {2} \\ \hline\end{array} This means that $$ \lim _{x \rightarrow 5^+} \frac{2 x-10}{x-5}=2 $$ Hence $$ \lim _{x \rightarrow 5 } \frac{2 x-10}{x-5}=2 $$
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