# How do you evaluate the limit (x^2-x-6)/(x+2) as x approaches -2?

Oct 29, 2016

You may do the division or use L'Hopital's Rule both tell you that it goes to -5.

#### Explanation:

${\lim}_{x \to - 2} \frac{{x}^{2} - x - 6}{x + 2} =$

Do the division:

${\lim}_{x \to - 2} x - 3 = - 5$

Or use L'Hopital's rule:

${\lim}_{x \to - 2} \frac{2 x - 1}{1} = - 5$

Oct 29, 2016

First of all...

$\frac{{x}^{2} - x - 6}{x + 2}$

$= \frac{\left(x + 2\right) \left(x - 3\right)}{x + 2}$

$= x - 3$

Now, in the limit as x approaches minus 2, x-3 will approach -5.