Chapter 14.1, Problem 12AYU

$\underset{x\to 3}{\lim }\frac{x^2-9}{x^2-\text{ 3}x}$

To determine

To find: The limit of $\underset{x\to 3}{\lim }\frac{\left(x^2-9\right)}{\left(x^2-3x\right)}$.

Answer to Problem 12AYU

Solution:

$2$

Explanation of Solution

Given:

$\underset{x\to 3}{\lim }\frac{\left(x^2-9\right)}{\left(x^2-3x\right)}$

Calculation:

$\underset{x\to 3}{\lim }\frac{\left(x^2-9\right)}{\left(x^2-3x\right)}=\underset{x\to 3}{\lim }\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}=\underset{x\to 3}{\lim }\left(\frac{x+3}{x}\right)=\frac{3+3}{3}=\frac{6}{3}=2$