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.
 A: 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 $x\to -\infty$.
A: Turning your second expression into your third isn't always too helpful for calculating the limit. In general, when you have the expression 
$$
\lim_{n \to  \infty}\frac{p(x)}{q(x)}
$$
where $p$ and $q$ are polynomials of the same degree, this limit is the ratio of the leading coefficients. In your example, this is 1.
A: I always tell my students to not divide anything and look at the powers.
If the highest power in the denominator is larger than that of the numerator, the limit at infinity is zero. If the highest power in the denominator is equal to that of the numerator, the limit at infinity is the ratio of the coefficients of those highest powers.
Here, the limiting behavior is $\frac{x^3}{|x|^3}$. The coefficients are $1$, so the limit is $1$.
A: 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
$$0 = \lim\limits_{x\to-\infty}(2/x^3) = \lim\limits_{x\to\infty}(2/x^3) = \lim\limits_{x\to-\infty}(1/x^3) = \lim\limits_{x\to\infty}(1/x^3)$$
?
