# How do you find vertical, horizontal and oblique asymptotes for [(e^-x)(x^5) + 2] /[ x^5 - x^4 -x +1]?

Jun 3, 2017

Define

$g \left(x\right) = \frac{{x}^{5} \cdot {e}^{-} x + 2}{{x}^{5} - {x}^{4} - x + 1}$

Bottom quintic should be factorised in order to find potential vertical asymptotes.

$f \left(x\right) = {x}^{5} - {x}^{4} - x + 1$

Can be seen easily that 1 and -1 are factors. Then $\left(x - 1\right) \left(x + 1\right)$, ${x}^{2} - 1$ is a factor of $f \left(x\right) = 0$. Then,

$f \left(x\right) = \left({x}^{2} - 1\right) \cdot \left(A {x}^{3} + B {x}^{2} + C x + D\right)$,
$f \left(x\right) = A {x}^{5} + B {x}^{4} + \left(C - A\right) {x}^{3} + \left(D - B\right) {x}^{2} - C x - D$

Comparing coefficients gives,

$f \left(x\right) = \left({x}^{2} - 1\right) \left({x}^{3} - {x}^{2} + x - 1\right)$.

Another obvious factor of the cubic is -1. Applying the same method,

$f \left(x\right) = {\left(x - 1\right)}^{2} \left(x + 1\right) \left({x}^{2} + 1\right)$.

Then we see,

$g \left(x\right) = \frac{{x}^{5} \cdot {e}^{-} x + 2}{{\left(x - 1\right)}^{2} \left(x + 1\right) \left({x}^{2} + 1\right)}$.

From this, we can deduce there are horizontal asymptotes as the denominator goes to 0.

Vertical asymptotes at $x = - 1$ and $x = 1$.

Oblique and horizontal asymptotes arise as $x$ goes to $\setminus \pm \setminus \infty$.

Rewrite $g \left(x\right)$ as,

$g \left(x\right) = \frac{{x}^{5}}{{\left(x - 1\right)}^{2} \left(x + 1\right) \left({x}^{2} + 1\right)} \cdot {e}^{-} x + \frac{2}{{\left(x - 1\right)}^{2} \left(x + 1\right) \left({x}^{2} + 1\right)}$.

Because the numerator and denominator are 5th order polynomials,
$\setminus {\lim}_{x \setminus \to + \setminus \infty} \left(\frac{{x}^{5}}{{\left(x - 1\right)}^{2} \left(x + 1\right) \left({x}^{2} + 1\right)}\right) = 1$.

Due to the nature of exponentials,
$\setminus {\lim}_{x \setminus \to + \setminus \infty} {e}^{-} x = 0$.

Clearly,
$\setminus {\lim}_{x \setminus \to + \setminus \infty} \left(\frac{2}{{\left(x - 1\right)}^{2} \left(x + 1\right) \left({x}^{2} + 1\right)}\right) = 0$.

Then,
${\lim}_{x \setminus \to + \setminus \infty} g \left(x\right) = 1 \cdot 0 + 0$
${\lim}_{x \setminus \to + \setminus \infty} g \left(x\right) = 0$

This gives us a horizontal asymptote $y = 0$.

Using the same limits, $g \left(x\right) \setminus \rightarrow {e}^{-} x$ as $x \setminus \rightarrow \setminus \infty$.

This gives us an oblique asymptote $y = {e}^{-} x$.