# How do you evaluate the limit (2x+4)/(x^2-3x-10) as x approaches 2?

Oct 3, 2016

$- \frac{2}{3}$

#### Explanation:

Since this function is not indeterminate at x = 2, we can evaluate it by direct substitution.

${\lim}_{x \to 2} \frac{2 x + 4}{{x}^{2} - 3 x - 10} = \frac{2 \left(2\right) + 4}{{2}^{2} - 3 \left(2\right) - 10} = \frac{8}{- 12} = - \frac{2}{3}$

Oct 3, 2016

${\lim}_{x \rightarrow 2} \frac{2 x + 4}{{x}^{2} - 3 x - 10} = - \frac{2}{3}$ and ${\lim}_{x \rightarrow - 2} \frac{2 x + 4}{{x}^{2} - 3 x - 10} = - \frac{2}{7}$

#### Explanation:

As $x$ approaches $2$, the numerator goes to $8$ and the denominator goes to $4 - 6 - 10 = - 12$, so the limit is $\frac{8}{-} 12 = - \frac{2}{3}$

As $x$ approaches $- 2$, both the numerator and denominator go to $0$. Since boith are polynomials and $02$ is a zero, we can be sure that $x - \left(- 2\right)$ = x+2 is a foctor of both the numerator and the denominator.

We'll factor and reduce.

lim_(xrarr-2)(2x+4)/(x^2-3x-10)=lim_(xrarr-2)(2(x+2))/(x+2)(x-5))#

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

$= \frac{2}{\left(- 2\right) - 5} = - \frac{2}{7}$