Precalculus (6th Edition) Blitzer

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

Chapter 11 - Cumulative Review Exercises - Page 1181: 18


The value of $\underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}$ is $\frac{5}{4}$.

Work Step by Step

Consider the expression: $\underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}$, To find the limit of the function, substitute $-3$ for x in above function, The function becomes, $\begin{align} & \underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}=\frac{{{\left( -3 \right)}^{2}}+\left( -3 \right)-6}{{{\left( -3 \right)}^{2}}+2\left( -3 \right)-3} \\ & =\frac{9-3-6}{9-6-3} \\ & =\frac{0}{0} \end{align}$ Here, the value is not defined at $x=-3$ Therefore, direct substitution is not possible. Therefore, $\begin{align} & \underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}=\underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+3x-2x-6}{{{x}^{2}}+3x-x-3} \\ & =\underset{x\to -3}{\mathop{\lim }}\,\frac{\left( {{x}^{2}}+3x \right)+\left( -2x-6 \right)}{\left( {{x}^{2}}+3x \right)+\left( -x-3 \right)} \\ & =\underset{x\to -3}{\mathop{\lim }}\,\frac{x\left( x+3 \right)-2\left( x+3 \right)}{x\left( x+3 \right)-\left( x+3 \right)} \\ & =\underset{x\to -3}{\mathop{\lim }}\,\frac{\left( x+3 \right)\left( x-2 \right)}{\left( x+3 \right)\left( x-1 \right)} \end{align}$ Eliminate the common terms, $\underset{x\to -3}{\mathop{\lim }}\,\frac{\left( x+3 \right)\left( x-2 \right)}{\left( x+3 \right)\left( x-1 \right)}=\underset{x\to -3}{\mathop{\lim }}\,\frac{x-2}{x-1}$ Now substitute $x=-3$ in the above function as shown below, $\begin{align} & \underset{x\to -3}{\mathop{\lim }}\,\frac{x-2}{x-1}=\frac{-3-2}{-3-1} \\ & =\frac{-5}{-4} \\ & =\frac{5}{4} \end{align}$ Therefore, the value of $\underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}$ is $\frac{5}{4}$.
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