# What are the asymptotes of f(x)=-x/((x-1)(x^2-4)) ?

Jul 22, 2018

$x = - 2$, x=1} and x=2$a r e v e r t i c a l a s y m p \to t e s o f$f(x). See explanations below.

#### Explanation:

Observ your function $f \left(x\right) = - \frac{x}{\left(x - 1\right) \left({x}^{2} - 4\right)}$
Asymptotes ( if they exist ) appear on a graph when ${\lim}_{x \to n} f \left(x\right) = \infty$, $n \in \mathbb{R}$, or if ${\lim}_{x \to \infty} f \left(x\right) = a x + b$, $\left(a , b\right) \in {\mathbb{R}}^{2}$

Here, $x \to 1$, $x \to 2$, $x \to - 2$ and $x \to \infty$ are good candidates.
( by product, we can see that $\left(x - 1\right) \left({x}^{2} - 4\right) \to 0$ for $x \to 1$, $x \to 2$, etc. )
So :
lim_(x to 1)f(x)=lim_(x to 1)-x/((x-1)(x²-4))
Let $X = x - 1 \iff x = X + 1$, with $X \to 0$ we can apply basic rules that we know on limits.
Now we have lim_(X to 0)-(X+1)/(X((X+1)²-4)

=lim_(X to 0)-(X+1)/(X(X²+2X-3)

=lim_(X to 0)-(X+1)/(X^3+2X^2-3X

Because we have the divison of polynomials, we only look at the mononial of smallest degree , here, $1$ and $3 X$.

So : ${\lim}_{x \to 1} f \left(x\right) = {\lim}_{X \to 0} f \left(X\right) = {\lim}_{X \to 0} \frac{1}{3 X} = \pm \infty$

Now let do the exact same thing for $x \to 2$ and $x \to - 2$

${\lim}_{x \to 2} f \left(x\right) = {\lim}_{x \to 2} - \frac{x}{\left(x - 1\right) \left({x}^{2} - 4\right)}$
Let $X = x - 2$

=lim_(X to 0)-(X+2)/((X+1)((X+2)²-4)

=lim_(X to 0)-(X+2)/((X+1)(X²+4X))

$= {\lim}_{X \to 0} - \frac{2}{4 X}$
$= \pm \infty$

I will not detail for $x \to - 2$, we find ${\lim}_{X \to 0} \frac{1}{6 X} = \pm \infty$

Now let see ${\lim}_{x \to \infty} f \left(x\right)$

=lim_(x to oo)-x/((x-1)(x²-4))=lim_(x to oo)-x/(x^3-x^2-4x+4)#
Now $x \to \infty$, so we only look at the monomials of highest degree.

$= {\lim}_{x \to \infty} - \frac{x}{{x}^{3}} = {\lim}_{x \to \infty} - \frac{1}{x} ^ 2 = {0}^{-}$

So they are no asymptotes in $\infty$.

We can easily see it on a graph :
graph{y=-x/((x-1)(x^2-4)) [-4,4,-8,8]}