Precalculus (6th Edition) Blitzer

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

Chapter 11 - Section 11.1 - Finding Limits Using Tables and Graphs - Concept and Vocabulary Check - Page 1137: 5

Answer

The complete statement is, “The limit notation $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L $ is called the right-hand limit. The notation means that as x gets closer to $a$ but remains greater than $a$, the corresponding value of $ f\left( x \right)$ gets closer to L.”

Work Step by Step

Consider the provided limit notation, $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L $. Here, $ f $ is any function defined on some open interval containing the number $ a $. The function f may or may not be defined at a. Hence, the notation $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L $ means that as $x$ gets closer to $a$ from the right, but remains unequal to $a$, the corresponding value of $ f\left( x \right)$ gets closer to L. Consider the limit notation, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)=6$ To solve the notation $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)=6$, substitute $ x=2$ in the limit notation $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)=6$. Therefore, $\begin{align} \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\ \left( {{2}^{2}}+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\ \left( 4+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\ 5.9\approx 6 & \\ \end{align}$ Thus, it is clear that, the limit notation $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L $ is called the right-hand limit. The notation means that as x gets closer to $a$ but remains greater than a, the corresponding value of $ f\left( x \right)$ gets closer to L.
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