Evaluating $\lim_{x\to -\infty} \frac{4x^3+1}{2x^3 + \sqrt{16x^6+1}}$ In finding this limit:
$$\lim_{x\to -\infty} \frac{4x^3+1}{2x^3 + \sqrt{16x^6+1}}$$ 
I've been told to divide all the terms by $-x^3$ (as opposed to $x^3$ if we take the limit as $x \to \infty$), and go from there. Dividing by a negative $x^3$ doesn't make sense to me, because we will be plugging in negative numbers approaching $-\infty$ anyways. Why double up? 
Is there a different way to think about/solve the limit?

You can do it either way.  If you divide by $x^3$ you have to remember that $x^3$ is a negative number and $x^3 =-\sqrt{x^6}$. you get
$\frac {4+\frac 1{x^3}}{2 +\frac {\sqrt{16x^6 + 1}}{x^3}}=$
$\frac {4+\frac 1{x^3}}{2+\frac {\sqrt{16x^6+1}}{-\sqrt{x^6}} }=$
$\frac {4+\frac 1{x^3}}{2-\sqrt{16+\frac 1{x^6}}}$
It's easier to avoid mistakes if you divide by $-x^3$ and get:
$\frac {-4-\frac 1{x^3}}{-2+\sqrt{\frac {16x^3+1}{(-x)^6}}}=\frac {-4-\frac 1{x^3}}{-2+\sqrt{16+\frac 1{x^6}}}$
