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: 17


The value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{x+10}-\frac{1}{10}}{x}$ is $\frac{-1}{100}$.

Work Step by Step

Consider the function $f\left( x \right)=\frac{\frac{1}{x+10}-\frac{1}{10}}{x}$ , Simplify the above function, $\begin{align} & f\left( x \right)=\frac{\frac{1}{x+10}-\frac{1}{10}}{x} \\ & =\frac{\frac{10-\left( x+10 \right)}{10\left( x+10 \right)}}{x} \\ & =\frac{\frac{10-x-10}{10\left( x+10 \right)}}{x} \\ & =\frac{-x}{10x\left( x+10 \right)} \end{align}$ $=\frac{-1}{10\left( x+10 \right)}$ Since, the function $g\left( x \right)=x+10$ is a polynomial, Find the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{x+10}-\frac{1}{10}}{x}$ , $\begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{x+10}-\frac{1}{10}}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{-1}{10\left( x+10 \right)} \\ & =\frac{-1}{10\left( \underset{x\to 0}{\mathop{\lim }}\,x+10 \right)} \\ & =\frac{-1}{10\left( 0+10 \right)} \\ & =\frac{-1}{10\left( 10 \right)} \end{align}$ $=\frac{-1}{100}$ Thus, the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{x+10}-\frac{1}{10}}{x}$ is $\frac{-1}{100}$.
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