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 - Exercise Set - Page 1138: 7

Answer

See below:

Work Step by Step

Consider the provided limit notation, $\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{x-2}$ . In making the table, choose the value of x close to 3 from the left and from the right as x approaches 3. As x approaches 3 from the left, arbitrarily start with $ x=2.99$. Then select two additional values of x that are closer to 3, but still less than 3; we choose 2.999 and 2.9999. Evaluate f at each chosen value of x to obtain the corresponding values of $ f\left( x \right)$. At $ x=2.99$, Substitute, $ x=2.99$ in the provided limit notation, $\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{x-2}$. Therefore, $\begin{align} & \underset{x\to 2.99}{\mathop{\lim }}\,\frac{1}{x-2}=\frac{1}{2.99-2} \\ & =\frac{1}{0.99} \\ & =1.0101 \end{align}$ Now, at $ x=2.999$ Substitute, $ x=2.999$ in the provided limit notation, $\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{x-2}$. $\begin{align} & \underset{x\to 2.999}{\mathop{\lim }}\,\frac{1}{x-2}=\frac{1}{2.999-2} \\ & =\frac{1}{0.999} \\ & =1.0010 \end{align}$ At $ x=2.9999$ Substitute, $ x=2.9999$ in the provided limit notation, $\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{x-2}$. $\begin{align} & \underset{x\to 2.9999}{\mathop{\lim }}\,\frac{1}{x-2}=\frac{1}{2.9999-2} \\ & =\frac{1}{0.9999} \\ & =1.0001 \end{align}$ Now, as x approaches 3 from the right, arbitrarily start with $ x=3.01$. Then select two additional values of x that are closer to 3, but still greater than 3; we choose 3.001 and 3.0001. At $ x=3.01$ Substitute, $ x=3.01$ in the provided limit notation, $\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{x-2}$ Therefore, $\begin{align} & \underset{x\to 3.01}{\mathop{\lim }}\,\frac{1}{x-2}=\frac{1}{3.01-2} \\ & =\frac{1}{1.01} \\ & =0.9901 \end{align}$ Now, at $ x=3.001$ Substitute, $ x=3.001$ in the provided limit notation, $\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{x-2}$ Therefore, $\begin{align} & \underset{x\to 3.001}{\mathop{\lim }}\,\frac{1}{x-2}=\frac{1}{3.001-2} \\ & =\frac{1}{1.001} \\ & =0.9990 \end{align}$ At $ x=3.0001$ Substitute, $ x=3.0001$ in the provided limit notation, $\underset{x\to 3}{\mathop{\lim }}\,\frac{1}{x-2}$ $\begin{align} & \underset{x\to 3.0001}{\mathop{\lim }}\,\frac{1}{x-2}=\frac{1}{3.0001-2} \\ & =\frac{1}{1.0001} \\ & =0.9999 \end{align}$ The limit of $\frac{1}{x-2}$ as x approaches 3 equals the number 1.
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