Why $x=\pm1$ is not an asymptote of $\frac{x^3}{x^2+1}$?

Question

By long division, $f\left(x\right)=\frac{x^3}{x^2+1}$is equal to $x-\frac{x}{x^2+1}$. Therefore, there is an asymptote $y=x$. But why there is no an asymptote $x=\pm1$? How to determine whether the denominator is an asymptote?

Thank you for your attention

Answer 1

Denominator is a vertical asymptote when it's zero for some real value. $x^2+1=0$ has only complex solutions $x=\pm i$.

If you changed the denominator to $x^2-1$, then there would be a vertical asymptotes at $\pm1$.

Answer 2

An asymptote is a restriction of an algebraic equation by means of geometry. For instance, say you have $R(x) = \frac{f(x)}{g(x)} $. Then we know that $g(x) \neq 0$. In other words, all the values of $x$ that make $g(x) = 0 $ are asymptotes. In your case, we have

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

Notice by trivial inequality, $x^2 \geq 0 $ which implies $x^2 + 1 \geq 1 $. Hence the denominator is always non-zero, so we are safe. In particular,

$$ \{ x : g(x) = 0 \} = \varnothing $$

Answer 3

If you see the limits in $\pm \infty$ the result is $\pm\infty$ therefore there is not horizontal asymptotes. And for finite numbers $a$ the limit as $x\to a$ ever exists. Then there is not vertical asymptotes.