## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1154: 68

#### Answer

The limit $\underset{x\to 2}{\mathop{\lim }}\,\sqrt{5x-6}$ is 2. And the corresponding limit property is $\underset{x\to a}{\mathop{\lim }}\,\sqrt[n]{f\left( x \right)}=\sqrt[n]{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}=\sqrt[n]{L}$, where $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$ and $n$ is any integer greater than or equal to 2 provided that all roots represent real numbers.

#### Work Step by Step

Consider the given limit $\underset{x\to 2}{\mathop{\lim }}\,\sqrt{5x-6}$. First, find $\underset{x\to 2}{\mathop{\lim }}\,\left( 5x-6 \right)$. Use the limit properties $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ and $\underset{x\to a}{\mathop{\lim }}\,c=c$, where $c$ is a constant. \begin{align} & \underset{x\to 2}{\mathop{\lim }}\,\left( 5x-6 \right)=\underset{x\to 2}{\mathop{\lim }}\,5x-\underset{x\to 2}{\mathop{\lim }}\,6 \\ & =10-6 \\ & =4 \end{align} The limit that is required is calculated by taking this limit 4, and taking its square root. Thus, $\underset{x\to 2}{\mathop{\lim }}\,\sqrt{\left( 5x-6 \right)}=\sqrt{4}=2$ In limit notation, the corresponding limit property is If $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$ and $n$ is any integer greater than or equal to 2, then $\underset{x\to a}{\mathop{\lim }}\,\sqrt[n]{f\left( x \right)}=\sqrt[n]{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}=\sqrt[n]{L}$, provided that all roots represent real numbers. In words, the limit of the nth root of a function is found by taking the limit of the function and then taking the nth root of this limit.

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