# What is the limit of (sqrt(9x^6 - 6)) / (x^3 + 1) as x goes to negative infinity?

Oct 9, 2015

Here is another way to write up the solution.

#### Explanation:

$\sqrt{9 {x}^{6} - 6} = \sqrt{{x}^{6} \left(9 - \frac{6}{x} ^ 6\right)}$ for $x \ne 0$

$= \sqrt{{x}^{6}} \sqrt{9 - \frac{6}{x} ^ 6}$ for $x \ne 0$

$= \left\mid {x}^{3} \right\mid \sqrt{9 - \frac{6}{x} ^ 6}$ for $x \ne 0$

And ${x}^{3} + 1 =$ for $x \ne 0$

For $x < 0$, we have $\left\mid {x}^{3} \right\mid = - {x}^{3}$

As $x \rightarrow - \infty$, we are concerned with negative values far from 0.

Putting all this together we write:

${\lim}_{x \rightarrow - \infty} \frac{\sqrt{9 {x}^{6} - 6}}{{x}^{3} + 1} = {\lim}_{x \rightarrow - \infty} \frac{\left\mid {x}^{3} \right\mid \sqrt{9 - \frac{6}{x} ^ 6}}{{x}^{3} \left(1 + \frac{1}{x} ^ 3\right)}$

$= {\lim}_{x \rightarrow - \infty} \frac{- {x}^{3} \sqrt{9 - \frac{6}{x} ^ 6}}{{x}^{3} \left(1 + \frac{1}{x} ^ 3\right)}$

$= {\lim}_{x \rightarrow - \infty} \frac{- \sqrt{9 - \frac{6}{x} ^ 6}}{1 + \frac{1}{x} ^ 3}$

$= \frac{- \sqrt{9 - 0}}{1 + 0}$

$= - 3$