Why does the modulus of $x$ become $-x$ as $x$ tends to negative infinity in this example? I came across this example in Thomas' calculus:
For $x>0$:
$$
\lim_{x\to -\infty} \frac{x^3-2}{|x|^3+1}=\lim_{x\to -\infty}\frac{x^3-2}{(-x)^3+1}
$$
I was always taught that the modulus of $x$ will always be positive, but here it has been made negative and I am not sure as to what the mathematical reasoning behind doing this is. Any help would be greatly appreciated.
(the full example is on page 105 of Thomas'Calculus: early transcendentals)
 A: The part “for $x>0$” makes no sense, if you want to compute the limit for $x\to-\infty$.
But, since you're interested in this limit, it is not restrictive to assume $x<0$, so $|x|=-x$ (which is indeed positive).
A: If you wanna know how limit is negative then let me tell you that for $x<0$ then $|x|=-x$

 Proof - $|x|$ will result in $x$ and $-x$. However since $x<0$, $x$ case is eliminated and so $|x|=-x$

A: When computing a limit of the form $\lim_{x \to -\infty}f(x)$, you can assume that $x<0$. This is because, by definition, $\lim_{x \to -\infty}f(x)=l$ means

For every $\varepsilon>0$ there is an $N>0$ such that $x<-N \implies|f(x)-l|<\varepsilon$.

Since $N>0$, we have $-N<0$, and so $x<-N<0$. Thus, we are only interested in the behaviour of $f(x)$ when $x$ is negative. So to compute
$$
\lim_{x \to -\infty}\frac{x^3-2}{|x|^3+1}
$$
we can assume that $x<0$, and so $|x|=-x$ (remembering that $-x$ is positive in this case).
As Egreg has already noted, it doesn't make sense to say

For $\color{red}{x>0}$:
$$
\lim_{x\to -\infty} \frac{x^3-2}{|x|^3+1}=\lim_{x\to -\infty}\frac{x^3-2}{(-x)^3+1}
$$

and so I'm assuming that you meant to say "for $\color{red}{x<0}$" instead.
