## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1155: 83

#### Answer

The value of the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}$ is $\frac{2}{3}$.

#### Work Step by Step

Consider the limit to be solved, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}$ Simplify it as follows: $\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x}+\frac{1}{3}\times \frac{\cos x-1}{x} \right)$ Using $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x}+\frac{1}{3}\times \frac{\cos x-1}{x} \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x} \right)+\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{3}\times \frac{\cos x-1}{x} \right) \end{align} Using $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)\times g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\times \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x}+\frac{1}{3}\times \frac{\cos x-1}{x} \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x} \right)+\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{3}\times \frac{\cos x-1}{x} \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,\frac{2}{3}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}+\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{3}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x-1}{x} \end{align} Using the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1\ \text{and }\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x-1}{x}=0$, also using limit property $\underset{x\to a}{\mathop{\lim }}\,c=c,c\text{ is constant}$. \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2}{3}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}+\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{3}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x-1}{x} \\ & =\frac{2}{3}\times 1+\frac{1}{3}\times 0 \\ & =\frac{2}{3}+0 \\ & =\frac{2}{3} \end{align} Thus, the value of the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}$ is $\frac{2}{3}$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.