# Horizontal Asymptote

• Mar 17th 2008, 12:52 PM
kdogg121
Horizontal Asymptote
Ok, we learned limits a while back so I am trying to scrape away the cobwebs, I am having trouble finding the horizontal asymptote to a function to which I eventually will need to sketch a curve. My professor said to use the power rule just like we used to evaluate limits, by multiplying by the reciprocal, so I multiplied each term by (1/X) but I cant get the value the book has for the asymptote. Here is the problem:

F(X)= 3X-1 / X+2

Book has answer of Y=3 for horizontal, can someone illustrate how I use the reciprocal power rule for this, thanks!!
• Mar 17th 2008, 01:00 PM
galactus
Something to take note of is when you are taking the lmit of a rational expression, everything but the highest powers can be disregarded.

This leaves you with $\frac{3x}{x}=3$

Also, just divide and you can see it right off.

$\frac{3x-1}{x+2}=3-\frac{7}{x+2}$

$\lim_{x\rightarrow{\infty}}\left[3-\frac{7}{x+2}\right]=?$

What your professor is probably talking about is dividing the top and bottom by x.

$\lim_{x\rightarrow{\infty}}\frac{\frac{3x}{x}-\frac{2}{x}}{\frac{x}{x}+\frac{2}{x}}=$

$\lim_{x\rightarrow{\infty}}\frac{3-\frac{1}{x}}{1+\frac{2}{x}}$

As you can see, the terms with the x in the denominator tend to 0 and you are left with 3.
• Mar 17th 2008, 01:17 PM
kdogg121

X^2 + 2 / X^2+1

I used the reciprocal of 1/X^2 and got

1+(2/X^2) / 1+X^2 but I don't find where this yields the asymptote of Y=1 as is the answer. What am I doing wrong?
• Mar 17th 2008, 01:34 PM
galactus
You are correct, it's right in front of you.

$\frac{\frac{x^{2}}{x^{2}}+\frac{2}{x^{2}}}{\frac{x ^{2}}{x^{2}}+\frac{1}{x^{2}}}$

$\frac{1+\frac{2}{x^{2}}}{1+\frac{1}{x^{2}}}$

Same as before, the terms with x^2 in the denominator tend to 0 as x-->infinity and you are left with 1 as the limit.
• Mar 17th 2008, 01:54 PM
kdogg121
So looking at both examples it looks like I just need to recognize that whatever constant term I get will be my limit? It seems like this is almost too easy. This looks totally different than the way we did it in class, but I guess if it works. Are we still following the reciprocal power rules method to obtain the asymptote?
• Mar 17th 2008, 02:07 PM
galactus
rational expressions are among the easiest to find limits.

Disregard everything but the terms with the highest powers.

If the denominator has the highest terms, then its limit is 0.

If the powers are the same, then the limit is the ratio of the leading coefficients.

If the numerator has the highest power, then the limit is infinity.
• Mar 17th 2008, 02:16 PM
kdogg121
Quote:

Originally Posted by galactus
rational expressions are among the easiest to find limits.

Disregard everything but the terms with the highest powers.

If the denominator has the highest terms, then its limit is 0.

If the powers are the same, then the limit is the ratio of the leading coefficients.

If the numerator has the highest power, then the limit is infinity.

I am not looking for limits though, just the asymptote, but your saying just look at the term in the function that has the highest power and divide the coefficients and that will be my asymptote?
• Mar 17th 2008, 02:30 PM
galactus
That is how you find a horizontal asymptote. By taking

$\lim_{x\rightarrow{\infty}}f(x)=L$

or

$\lim_{x\rightarrow{-\infty}}f(x)=L$

That's what an asymptote is.