Question

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

Answer 1

Define

`g(x)=(x^5*e^-x+2)/(x^5-x^4-x+1)`

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

`f(x)=x^5-x^4-x+1`

Can be seen easily that 1 and -1 are factors. Then `(x-1)(x+1)`, `x^2-1` is a factor of `f(x)=0`. Then,

`f(x) = (x^2-1)*(Ax^3+Bx^2+Cx+D)`,
`f(x)=Ax^5 +Bx^4 + (C-A)x^3+(D-B)x^2-Cx-D`

Comparing coefficients gives,

`f(x)=(x^2-1)(x^3-x^2+x-1)`.

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

`f(x)=(x-1)^2(x+1)(x^2+1)`.

Then we see,

`g(x)=(x^5*e^-x+2)/((x-1)^2(x+1)(x^2+1))`.

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 `\pm\infty`.

Rewrite `g(x)` as,

`g(x) = (x^5)/((x-1)^2(x+1)(x^2+1))*e^-x+(2)/((x-1)^2(x+1)(x^2+1))`.

Because the numerator and denominator are 5th order polynomials,
`\lim_{x\to+\infty} ((x^5)/((x-1)^2(x+1)(x^2+1))) = 1`.

Due to the nature of exponentials,
`\lim_{x\to+\infty} e^-x = 0`.

Clearly,
`\lim_{x\to+\infty} ((2)/((x-1)^2(x+1)(x^2+1))) = 0`.

Then,
`lim_{x\to+\infty}g(x)=1*0+0`
`lim_{x\to+\infty}g(x)=0`

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

Using the same limits, `g(x) \rightarrow e^-x` as `x \rightarrow \infty`.

This gives us an oblique asymptote `y=e^-x`.