Determining asymptote Question

H

hank

Junior Member
Joined
Sep 13, 2006
Messages
209
I have this problem where I have to determine the asymptotes for f(x) = (x - 2)^3 / x^2.

The horizontal asymptote is easy enough, x = 0.

However, the problem ends up with an oblique asymptote, y = x - 6, and I'm not sure how to get this.

I utilized limits to get the following:
lim x->inf f(x) = x +6, and
lim x->-inf f(x) = -x - 6.

I'm not sure how to translate the limits into the asymptote y = x - 6. Can someone explain that?
 
Actually, I think I got it.

When I was figuring the limits, I made a mistake on expanding the (x-2)^3 binomial.
So the first limit is x-6 and the 2d limit is -x - 6, which when you plug in a negative number for x becomes x - 6.

So they both match and that's the limit.

Am I right?
 
hank said:
... So the first limit is x-6 and the [2nd] limit is -x - 6, which when you plug in a negative number for x becomes x - 6.

So they both match and that's the limit.

Am I right?

No.
Click to expand...


Hello Hank:

Whenever a limit exists, it always equals some real number (not an expression containing a variable).

Therefore, it is not correct to say that a limit equals x - 6 or -x - 6.

\(\displaystyle \lim_{x \to \infty} \frac{(x - 2)^3}{x^2} \; = \; \infty\)

\(\displaystyle \lim_{x \to -\infty} \frac{(x - 2)^3}{x^2} \; = \; -\infty\)

Infinity is not a number; therefore, these limits do not exist. We write these two limit statements to indicate the behavior of f as x increases (or decreases) without bound.

These two statements tell us that f does not approach any fixed value when x becomes huge positively or negatively. The limits do not exist.

Regarding any asymptotes (other than the vertical one that you already found), here is how I would proceed.

First, I would recognize that, since the degree of the polynomial in the numerator is one more than the degree of the polynomial in the denominator, the rational function f has a slant asymptote.

To find the equation of a slant asymptote, I would normally carry out longhand polynomial division in order to get a quotient and remainder. This first requires expanding the factored polynomial in the numerator. However, since the denominator in this function is a monomial (a single term), there is no need for the longhand division.

\(\displaystyle \frac{x^3 - 6x^2 + 12x - 8}{x^2} \; = \; \frac{x^3}{x^2} - \frac{6x^2}{x^2} + \frac{12x}{x^2} - \frac{8}{x^2} \;= \; x - 6 + \frac{12}{x} - \frac{8}{x^2}\)

It's clear that the terms 12/x and -8/x^2 go to zero as x increases (or decreases) without bound.

That leaves y = x - 6 for the equation of the slant asymptote.

Cheers,

~ Mark :)

 
You must log in or register to reply here.
Top