# 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?

Here's a slight variation: you can reflect the variable, so that it approaches $$\infty$$. Let $$y = -x$$. Then, as $$x \to -\infty$$, $$y \to \infty$$, and we get $$\lim_{y \to \infty} \frac{-4y^3 + 1}{-2y^3 + \sqrt{16y^6 + 1}}.$$ Now you can divide top and bottom by $$y^3$$.

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:

$$\frac {-4-\frac 1{x^3}}{-2+\sqrt{\frac {16x^3+1}{(-x)^6}}}=\frac {-4-\frac 1{x^3}}{-2+\sqrt{16+\frac 1{x^6}}}$$

The given hint does make sense.

If $$x<0$$ then $$0<-x^3=\sqrt{x^6}$$, and by dividing all terms by $$-x^3$$, we get $$\frac{-4-\frac{1}{x^3}}{-2 + \sqrt{16+\frac{1}{x^6}}}.$$ Now it should be quite easy to find the limit as $$x\to -\infty$$?

• @Buddhapus I edited my answer. Is it clear now? Sep 5, 2019 at 5:57