# How do you evaluate each of the following limits, if it exists lim (3x^2+5x-2)/(x^2-3x-10) as x-> -2?

Dec 14, 2016

${\lim}_{x \rightarrow - 2} \frac{3 {x}^{2} + 5 x - 2}{{x}^{2} - 3 x - 10} = 1$

#### Explanation:

To do this, factor the numerators and denominators. Notice that we can't plug in $- 2$ now, because this gives us the ratio $0 / 0$.

To factor $3 {x}^{2} + 5 x - 2$, we look for two numbers whose product is $- 6$ and sum is $5$, which are $6$ and $- 1$.

Then, $3 {x}^{2} + 5 x - 2 = 3 {x}^{2} + 6 x - x - 2$

$\textcolor{w h i t e}{l l l} = 3 x \left(x + 2\right) - 1 \left(x + 2\right)$

$\textcolor{w h i t e}{l l l} = \left(3 x - 1\right) \left(x + 2\right)$

In factoring ${x}^{2} - 3 x - 10$, we look for the numbers whose product is $- 10$ and sum is $- 3$, which are $- 5$ and $2$.

Then, ${x}^{2} - 3 x - 10 = \left(x - 5\right) \left(x + 2\right)$

So ${\lim}_{x \rightarrow - 2} \frac{3 {x}^{2} + 5 x - 2}{{x}^{2} - 3 x - 10} = {\lim}_{x \rightarrow - 2} \frac{\left(3 x - 1\right) \left(x + 2\right)}{\left(x - 5\right) \left(x + 2\right)}$, and we see that we can cancel the $\left(x + 2\right)$ terms, so the limit equals ${\lim}_{x \rightarrow - 2} \frac{3 x - 1}{x - 5}$.

Now plugging in $- 2$ won't create the issue it did before and we can plug in $x = - 2$, so: ${\lim}_{x \rightarrow - 2} \frac{3 x - 1}{x - 5} = \frac{3 \left(- 2\right) - 1}{- 2 - 5} = \frac{- 7}{- 7} = 1$.