# How do you determine the limit of 1/(x²+5x-6) as x approaches -6?

Apr 21, 2016

DNE-does not exist

#### Explanation:

${\lim}_{x \to - 6} \frac{1}{\left(x + 6\right) \left(x - 1\right)}$

$= \frac{1}{0 \cdot - 7}$

$= \frac{1}{0}$

$D N E$

Apr 21, 2016

The limit does not exist. Look at the signs of the factors.

#### Explanation:

Let $f \left(x\right) = \frac{1}{{x}^{2} + 5 x - 6} = \frac{1}{\left(x + 6\right) \left(x - 1\right)}$

Not that as $x \rightarrow - 6$, we have $\left(x - 1\right) \rightarrow - 7$

From the left

As $x \rightarrow - {6}^{-}$, the factor $\left(x + 6\right) \rightarrow {0}^{-}$, so $f \left(x\right)$ is positive and increasing without bound.

${\lim}_{x \rightarrow - {6}^{-}} f \left(x\right) = \infty$

From the right

As $x \rightarrow - {6}^{+}$, the factor $\left(x + 6\right) \rightarrow {0}^{+}$, so $f \left(x\right)$ is negative and increasing without bound.

${\lim}_{x \rightarrow - {6}^{+}} f \left(x\right) = - \infty$

Two sided

${\lim}_{x \rightarrow - 6} f \left(x\right)$ does not exist.