Chapter 14.2, Problem 29AYU

In Problems 7-42, find each limit algebraically.

$\underset{x\to 2}{\lim }\frac{x^2-4}{x^2-2x}$

To determine

To find: $\underset{x \to 2}{\text{lim}}\frac{x^2 - 4}{x^2 - 2x}$

Answer to Problem 29AYU

2

Explanation of Solution

Given:

$\underset{x \to 2}{\text{lim}}\frac{x^2 - 4}{x^2 - 2x}$

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 $f\left(x\right) = \underset{x \to 2}{\text{lim}}\frac{x^2 - 4}{x^2 - 2x}$, at $f\left(2\right)$, the value is $0/0$ which is undefined, so we have to simplify it.

so $\frac{x^2 - 4}{x^2 - 2x} = \frac{\left(x + 2\right)\left(x - 2\right)}{x\left(x - 2\right)} = \frac{x + 2}{x}$

Now $\underset{x \to 2}{\text{lim}}\frac{x^2 - 4}{x^2 - 2x} = \underset{x \to 2}{\text{lim}}\frac{x + 2}{x}$

$\underset{x \to 2}{\text{lim}}\frac{x + 2}{x} = \frac{\underset{x \to 2}{\text{lim}}x\text{ + 2}}{\underset{x \to 2}{\text{lim}}x}$

$= \frac{2 + 2}{2}$

$= \frac{4}{2}$

$= 2$

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