## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.1 - Finding Limits Using Tables and Graphs - Exercise Set - Page 1138: 6

#### Answer

See below: #### Work Step by Step

Consider the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$ . In making the table, choose the value of x close to 2 from the left and from the right as x approaches 2. As x approaches 2 from the left, arbitrarily start with $x=1.99$. Then select two additional values of x that are closer to 2, but still less than 2; we choose 1.999 and 1.9999. Evaluate f at each chosen value of x to obtain the corresponding values of $f\left( x \right)$. At $x=1.99$, Substitute, $x=1.99$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$. Therefore, \begin{align} & \underset{x\to 1.99}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)=\left( {{\left( 1.99 \right)}^{2}}-1 \right) \\ & =\left( 3.9601-1 \right) \\ & =2.9601 \end{align} Now, at $x=1.999$ Substitute, $x=1.999$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$. \begin{align} & \underset{x\to 1.999}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)=\left( {{\left( 1.999 \right)}^{2}}-1 \right) \\ & =\left( 3.9960-1 \right) \\ & =2.9960 \end{align} At $x=1.9999$ Substitute, $x=1.9999$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$. \begin{align} & \underset{x\to 1.9999}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)=\left( {{\left( 1.9999 \right)}^{2}}-1 \right) \\ & =\left( 3.9996-1 \right) \\ & =2.9996 \end{align} Now, as x approaches 2 from the right, arbitrarily start with $x=2.01$. Then select two additional values of x that are closer to 2, but still greater than 2; we choose 2.001 and 2.0001. At $x=2.01$ Substitute, $x=2.01$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$ \begin{align} & \underset{x\to 2.01}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)=\left( {{\left( 2.01 \right)}^{2}}-1 \right) \\ & =\left( 4.0401-1 \right) \\ & =3.0401 \end{align} Now, at $x=2.001$ Substitute, $x=2.001$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$ Therefore, \begin{align} & \underset{x\to 2.001}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)=\left( {{\left( 2.001 \right)}^{2}}-1 \right) \\ & =\left( 4.0040-1 \right) \\ & =3.0040 \end{align} At $x=2.0001$ Substitute, $x=2.0001$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$ \begin{align} & \underset{x\to 2.0001}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)=\left( {{\left( 2.0001 \right)}^{2}}-1 \right) \\ & =\left( 4.0004-1 \right) \\ & =3.0004 \end{align} The limit of $\left( {{x}^{2}}-1 \right)$ as x approaches 2 equals the number 3.

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