# How do you find the limit of (x^2-4)/(x-2) as x approaches 2?

Nov 3, 2016

${\lim}_{x \rightarrow 2} \frac{{x}^{2} - 4}{x - 2} = 4$

#### Explanation:

If we look at the graph of $y = \frac{{x}^{2} - 4}{x - 2}$ we can see that it is clear that the limit exists, and is approximately $4$

graph{(x^2-4)/(x-2) [-10, 10, -5, 5]}

The numerator is the difference of two squares, and as such we can factorise using it as

${A}^{2} - {B}^{2} \equiv \left(A + B\right) \left(A - B\right)$

Se we can factorise as follows:

${\lim}_{x \rightarrow 2} \frac{{x}^{2} - 4}{x - 2} = {\lim}_{x \rightarrow 2} \frac{{x}^{2} - {2}^{2}}{x - 2}$

$= {\lim}_{x \rightarrow 2} \frac{\left(x + 2\right) \left(x - 2\right)}{x - 2}$
$= {\lim}_{x \rightarrow 2} \frac{\left(x + 2\right) \cancel{x - 2}}{\cancel{x - 2}}$
$= {\lim}_{x \rightarrow 2} \left(x + 2\right)$
$= \left(2 + 2\right)$
$= 4$

Which is completely consistent with the above graph.