Chapter 14.2, Problem 30AYU

In Problems 7-42, find each limit algebraically.

$\underset{x\to -1}{\lim }\frac{x^2+x}{x^2-1}$

To determine

To find: $\underset{x \to -1}{\text{lim}}\frac{x^2 + x}{x^2 - 1}$

Answer to Problem 30AYU

$\frac{1}{2}$

Explanation of Solution

Given:

$\underset{x \to -1}{\text{lim}}\frac{x^2 + x}{x^2 - 1}$

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 $\underset{x \to c}{\text{lim}}g\left(x\right) \ne 0$

$a^2 - b^2 = \left(a + b\right)\left(a - b\right)$

Calculation:

For $f\left(x\right) = \underset{x \to -1}{\text{lim}}\frac{x^2 + x}{x^2 - 1}$, at $f\left(-1\right)$, the value is $0/0$ which is undefined, so we have to simplify it.

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

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

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

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

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

So $\underset{x \to -1}{\text{lim}}\frac{x^2 + x}{x^2 - 1}$ value is $\frac{1}{2}$.