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: 10

Answer

See below:
1571109494

Work Step by Step

Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}$. In making the table, choose the value of x close to 0 from the left and from the right as x approaches 0. As x approaches 0 from the left, arbitrarily start with $ x=-0.01$. Then select two additional values of x that are closer to 0, but still less than 0; we choose $-0.001$ and $-0.0001$. Evaluate f at each chosen value of x to obtain the corresponding values of $ f\left( x \right)$. At $ x=-0.01$, Substitute, $ x=-0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}$. Therefore, $\begin{align} & \underset{x\to -0.01}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}=\frac{-0.01+1}{{{\left( -0.01 \right)}^{2}}+1} \\ & =\frac{-0.01+1}{0.0001+1} \\ & =\frac{0.99}{1.0001} \\ & =0.9899 \end{align}$ Now, at $ x=-0.001$ Substitute, $ x=-0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}$. $\begin{align} & \underset{x\to -0.001}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}=\frac{-0.001+1}{{{\left( -0.001 \right)}^{2}}+1} \\ & =\frac{0.999}{0.000001+1} \\ & =\frac{0.999}{1.000001} \\ & =0.9990 \end{align}$ At $ x=-0.0001$ Substitute, $ x=-0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}$. $\begin{align} & \underset{x\to -0.0001}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}=\frac{-0.0001+1}{{{\left( -0.0001 \right)}^{2}}+1} \\ & =\frac{-0.0001+1}{0.00000001+1} \\ & =\frac{0.9999}{1.00000001} \\ & =0.9999 \end{align}$ Now, as x approaches 0 from the right, arbitrarily start with $ x=0.01$. Then select two additional values of x that are closer to 0, but still greater than 0; we choose 0.001 and 0.0001. At $ x=0.01$ Substitute, $ x=0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}$. Therefore, $\begin{align} & \underset{x\to 0.01}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}=\frac{0.01+1}{{{\left( 0.01 \right)}^{2}}+1} \\ & =\frac{0.01+1}{0.0001+1} \\ & =\frac{1.01}{1.0001} \\ & =1.0099 \end{align}$ Now, at $ x=0.001$ Substitute, $ x=0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}$. Therefore, $\begin{align} & \underset{x\to 0.001}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}=\frac{0.001+1}{{{\left( 0.001 \right)}^{2}}+1} \\ & =\frac{0.001+1}{0.000001+1} \\ & =\frac{1.001}{1.000001} \\ & =1.0010 \end{align}$ At $ x=0.0001$ Substitute, $ x=0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}$. $\begin{align} & \underset{x\to 0.0001}{\mathop{\lim }}\,\frac{x+1}{{{x}^{2}}+1}=\frac{0.0001+1}{{{\left( 0.0001 \right)}^{2}}+1} \\ & =\frac{0.0001+1}{0.00000001+1} \\ & =\frac{1.0001}{1.00000001} \\ & =1.0001 \end{align}$ The limit of $\frac{x+1}{{{x}^{2}}+1}$ as x approaches 0 equals the number 1.
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