Find the horizontal asymptotes of $f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$ Taken from Thomas' Calculus 12e
Find the horizontal asymptote of the graph of:

$$f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$$

Solution: We calculate the limits as ${x \to \pm \infty}$

For $x\ge0$: $$\lim\limits_{x\to\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to\infty}\frac{x^3-2}{x^3+1}=\lim\limits_{x\to\infty}\frac{1-(2/x^3)}{1+(1/x^3)}=1$$

I understand all except why the answer is one. I actually cannot find out how to get the answer.
As you find out, the limit as $x\to \infty$ is $1$ then $y=1$ is an horizontal asymptote.

For $x\to -\infty$ we have

$$\lim\limits_{x\to-\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to-\infty}\frac{x^3-2}{-x^3+1}=\lim\limits_{x\to-\infty}\frac{1-(2/x^3)}{-1+(1/x^3)}=-1$$

then $y=-1$ is an horizontal asymptote for $x\to -\infty$.
I assume the function's domain is $\mathbb R$ except where $f$ is not defined namely $x=-1$.

Now if $\lim_{x \to \infty} f(x) = a$ for some $a \in \mathbb R$, then $y=a$ is a horizontal asymptote.

Similarly, if $\lim_{x \to -\infty} f(x) = b$, for some $b \in \mathbb R$, then $y=b$ is a horizontal asymptote.

For $x\to\infty$, we have

$$\lim\limits_{x\to\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to\infty}\frac{x^3-2}{x^3+1}$$

$$=\lim\limits_{x\to\infty}\frac{x^3/x^3-(2/x^3)}{x^3/x^3+(1/x^3)}=\lim\limits_{x\to\infty}\frac{1-(2/x^3)}{1+(1/x^3)}$$

$$=\frac{\lim\limits_{x\to\infty} [1-(2/x^3)]}{\lim\limits_{x\to\infty} [1+(1/x^3)]}=\frac{ [\lim\limits_{x\to\infty}1-\lim\limits_{x\to\infty}(2/x^3)]}{ [\lim\limits_{x\to\infty}1+\lim\limits_{x\to\infty}(1/x^3)]}$$

$$=\frac{ [1-\lim\limits_{x\to\infty}(2/x^3)]}{ [1+\lim\limits_{x\to\infty}(1/x^3)]} = \frac{ [1-0]}{ [1+0]} = \frac11 = 1$$

Thus, $y=1$ is a vertical asymptote

Similarly, for $x\to-\infty$, we have

$$\lim\limits_{x\to-\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to-\infty}\frac{x^3-2}{(-x)^3+1}$$

$$= \cdots$$

$$= \cdots$$

$$= \frac{ [1-\lim\limits_{x\to-\infty}(2/x^3)]}{ [-1+\lim\limits_{x\to-\infty}(1/x^3)]} = \frac{ [1-0]}{ [-1+0]} = \frac1{-1} = -1$$

Thus, $y=-1$ is a vertical asymptote

Soooo...do you know why

$$0 = \lim\limits_{x\to-\infty}(2/x^3) = \lim\limits_{x\to\infty}(2/x^3) = \lim\limits_{x\to-\infty}(1/x^3) = \lim\limits_{x\to\infty}(1/x^3)$$

?