# Question #becac

Oct 24, 2016

${\lim}_{x \to - 1} \frac{\sqrt{{x}^{2} + 8} - 3}{x + 1} = - \frac{1}{3}$

#### Explanation:

As we are getting a $0$ in the numerator and the denominator, on direct substitution, our strategy will be to rationalize the numerator by using the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$. When we have two polynomial expressions with $- 1$ as a root, we will be able to cancel the $x + 1$ factors.

${\lim}_{x \to - 1} \frac{\sqrt{{x}^{2} + 8} - 3}{x + 1} = {\lim}_{x \to - 1} \frac{\left(\sqrt{{x}^{2} + 8} - 3\right) \left(\sqrt{{x}^{2} + 8} + 3\right)}{\left(x + 1\right) \left(\sqrt{{x}^{2} + 8} + 3\right)}$

$= {\lim}_{x \to - 1} \frac{{x}^{2} + 8 - 9}{\left(x + 1\right) \left(\sqrt{{x}^{2} + 8} + 3\right)}$

$= {\lim}_{x \to - 1} \frac{{x}^{2} - 1}{\left(x + 1\right) \left(\sqrt{{x}^{2} + 8} + 3\right)}$

$= {\lim}_{x \to - 1} \frac{\left(x + 1\right) \left(x - 1\right)}{\left(x + 1\right) \left(\sqrt{{x}^{2} + 8} + 3\right)}$

$= {\lim}_{x \to - 1} \frac{x - 1}{\sqrt{{x}^{2} + 8} + 3}$

$= \frac{- 1 - 1}{\sqrt{{\left(- 1\right)}^{2} + 8} + 3}$

$= - \frac{1}{3}$