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: 15


The value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-7x-15}{x-5}$ is $13$.

Work Step by Step

Consider the function $ f\left( x \right)=\frac{2{{x}^{2}}-7x-15}{x-5}$, Simplify the above function, $\begin{align} & f\left( x \right)=\frac{2{{x}^{2}}-7x-15}{x-5} \\ & =\frac{2{{x}^{2}}-10x+3x-15}{x-5} \\ & =\frac{2x\left( x-5 \right)+3\left( x-5 \right)}{x-5} \\ & =\frac{\left( x-5 \right)\left( 2x+3 \right)}{\left( x-5 \right)} \end{align}$ Further simplify the above expression, $\begin{align} & f\left( x \right)=\frac{\left( x-5 \right)\left( 2x+3 \right)}{\left( x-5 \right)} \\ & =2x+3 \end{align}$ The function $ f\left( x \right)=2x+3$ is a polynomial. Now, find the value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-7x-15}{x-5}$ $\begin{align} & \underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-7x-15}{x-5}=\underset{x\to 5}{\mathop{\lim }}\,2x+3 \\ & =2\underset{x\to 5}{\mathop{\lim }}\,x+3 \\ & =2\left( 5 \right)+3 \\ & =13 \end{align}$ Thus, the value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-7x-15}{x-5}$ is $13$.
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