Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 3 - The Derivative - 3.1 Limits - 3.1 Exercises - Page 137: 53

Answer

$$ f(x)=\left\{\begin{array}{ll}{x^{3}+2} & {\text { if } x \neq-1} \\ {5} & {\text { if } x=-1}\end{array}\right. $$ We obtain that: $$ \lim _{x \rightarrow-1^{-}} f(x)=1 $$ and $$ \lim _{x \rightarrow-1^{+}} f(x)=1 $$ Therefore $$ \lim _{x \rightarrow-1} f(x)=1 $$

Work Step by Step

$$ \text { Let } f(x)=\left\{\begin{array}{ll}{x^{3}+2} & {\text { if } x \neq-1} \\ {5} & {\text { if } x=-1}\end{array}\right. $$ A function $f(x)$ defined by two cases is called a piecewise function. The domain of $f(x)$ is all real numbers. To determine the limit as $x$ approaches -1, we are concerned only with the values of $f(x)$ when $x$ is close but not equal to -1 Once again, $$ \lim _{x \rightarrow-1^{-}} f(x)=1 $$ and $$ \lim _{x \rightarrow-1^{+}} f(x)=1 $$ Therefore $$ \lim _{x \rightarrow-1} f(x)=1 $$
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