Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Mid-Chapter Check Point - Page 1163: 3


The value of $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$ does not exist.

Work Step by Step

The value of $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$ exists only if the values of $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ and $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ exist and are equal. It is seen from the graph that, the value of $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $2$ and the value of $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $1$. Since $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, Thus, the value of $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$ does not exist.
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