Chapter 14.2, Problem 32AYU

In Problems 7-42, find each limit algebraically.

$\underset{x\to -3}{\lim }\frac{x^2+x-6}{x^2+2x-3}$

To determine

To find: $\underset{x \to -3}{\text{lim}}\frac{x^2 + x - 6}{x^2 + 2x - 3}$

Answer to Problem 32AYU

$\frac{5}{4}$

Explanation of Solution

Given:

$\underset{x \to -3}{\text{lim}}\frac{x^2 + x - 6}{x^2 + 2x - 3}$

Formula used:

$\underset{x \to c}{\text{lim}}\left[\frac{f\left(x\right)}{g\left(x\right)}\right] = \frac{\underset{x \to c}{\text{lim}}f\left(x\right)}{\underset{x \to c}{\text{lim}}g\left(x\right)}$, provided that $\underset{x \to c}{\text{lim}}g\left(x\right) \ne 0$

Calculation:

For $f\left(x\right) = \underset{x \to -3}{\text{lim}}\frac{x^2 + x - 6}{x^2 + 2x - 3}$, at $f(-3)$, the value is $0/0$ which is undefined, so we have to simplify it.

So $\frac{x^2 + x - 6}{x^2 + 2x - 3} = \frac{\left(x + 3\right)\left(x - 2\right)}{\left(x + 3\right)\left(x - 1\right)} = \frac{x - 2}{x - 1}$

Now $\underset{x \to -3}{\text{lim}}\frac{x^2 + x - 6}{x^2 + 2x - 3} = \underset{x \to -3}{\text{lim}}\frac{x - 2}{x - 1}$

$\underset{x \to -3}{\text{lim}}\frac{x - 2}{x - 1} = \frac{\underset{x \to -3}{\text{lim}}x - 2}{\underset{x \to -3}{\text{lim}}x - 1}$

$= \frac{-3 - 2}{-3 - 1}$

$= \frac{-5}{-4} = \frac{5}{4}$

So $\underset{x \to -3}{\text{lim}}\frac{x^2 + x - 6}{x^2 + 2x - 3}$ value is $\frac{5}{4}$.