# Question #84e30

Sep 7, 2016

As presented, the limit does not exist.

#### Explanation:

As $x \rightarrow - 3$, the numerator goes to $\sqrt{10}$ and the denominator goes to $0$, so the absolute value of the ratio is increasing without bound.

The limit as $x \rightarrow - {3}^{+}$ has the numerator approaching $\sqrt{10}$ while the denominator goes to $0$ through positive values. The ratio increases without bound.

That is ${\lim}_{x \rightarrow - {3}^{+}} \frac{\sqrt{{x}^{2} - x - 2} - 2}{x + 3} = \infty$

From the other side, as $x \rightarrow - {3}^{-}$ has the numerator approaching $\sqrt{10}$ while the denominator goes to $0$ through negative values. The ratio decreases without bound.

That is ${\lim}_{x \rightarrow - {3}^{-}} \frac{\sqrt{{x}^{2} - x - 2} - 2}{x + 3} = - \infty$.

The two-sided limit simply does not exist.

Sep 8, 2016

If the intended problem is ${\lim}_{x \rightarrow - {3}^{+}} \frac{\sqrt{{x}^{2} + x - 2} - 2}{x + 3}$, then the limit is $- \frac{5}{4}$

#### Explanation:

The initial form of ${\lim}_{x \rightarrow - {3}^{+}} \frac{\sqrt{{x}^{2} + x - 2} - 2}{x + 3}$ is the indeterminate $\frac{0}{0}$. Try rationalizing the numerator.
(You may not be sure it will work, but you need to try something. I am sure if will work because I've done lots of limits like this.)

${\lim}_{x \rightarrow - 3 \cdot +} \frac{\sqrt{{x}^{2} + x - 2} - 2}{x + 3} = {\lim}_{x \rightarrow - {3}^{+}} \frac{\left(\sqrt{{x}^{2} + x - 2} - 2\right)}{\left(x + 3\right)} \frac{\left(\sqrt{{x}^{2} + x - 2} + 2\right)}{\left(\sqrt{{x}^{2} + x - 2} + 2\right)}$

$= {\lim}_{x \rightarrow - {3}^{+}} \frac{{x}^{2} + x - 2 - 4}{\left(x + 3\right) \left(\sqrt{{x}^{2} + x - 2} + 2\right)}$

$= {\lim}_{x \rightarrow - {3}^{+}} \frac{{x}^{2} + x - 6}{\left(x + 3\right) \left(\sqrt{{x}^{2} + x - 2} + 2\right)}$

Note that we still get $\frac{0}{0}$, but we can factor and reduce now.

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

$= {\lim}_{x \rightarrow - {3}^{+}} \frac{x - 2}{\sqrt{{x}^{2} + x - 2} + 2}$

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

$= \frac{- 5}{\sqrt{4} + 2} = \frac{- 5}{4}$