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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?

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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 $y^3$.

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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:

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

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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 $x\to -\infty$?

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  • $\begingroup$ @Buddhapus I edited my answer. Is it clear now? $\endgroup$
    – Robert Z
    Sep 5, 2019 at 5:57

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