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.

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