# How do you find the limit of  (x^2 + 3x - 10) as x approaches 2^+?

May 16, 2016

${\lim}_{x \rightarrow {2}^{+}} \left({x}^{2} + 3 x - 10\right) = 0$.

#### Explanation:

As $x \rightarrow 2$ (from either side), ${x}^{2} \rightarrow 4$ and $3 x \rightarrow 6$, so ${\lim}_{x \rightarrow 2} \left({x}^{2} + 3 x - 10\right) = 4 + 6 - 10 = 0$.

It is possible that this is not all the information needed.

The limit is $0$, but in some situations we also need to know whether the values are approaching $0$ through positive or negative numbers.

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

As $x \rightarrow {2}^{+}$, both of these factors are positive, so we might write

${\lim}_{x \rightarrow 2} \left({x}^{2} + 3 x - 10\right) = {0}^{+}$.

We would need this if, for example, we wanted to evaluate

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

The numerator approaches $- 1$ and the denominator approaches $0$ but is positive, so the ratio is decreasing without bound. We write

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