Finding Asymptotes in Calculus: Rules & Explanation

[shooba]

When finding the asymptote of a rational function, if the degree of the numerator is less than or equal to the denominator you can divide by the highest power in the denominator and then take the limit as x goes to +/- infinity.
Apparently this trick does not work when the numerator is of higher degree and you are finding the slant asymptote, and you must use long division. I assume this has something to do with the fact that there isn't really a limit because you aren't approaching a number, but it seems like all the terms with x in their denominator should approach 0 and the x-to-the-something term in the numerator should dominate. I guess the terms with x in their denominator do not approach zero at the same "rate", but why is this not a problem when finding the horizontal asymptote?
I am in my first semester of calculus right now, if that limits the discussion somehow.

 

[Office_Shredder]

If p(x)=a_n x^n + a_{n-1} x^{n-1} +...+a_0 and q(x)=b_m x^m + b_{m-1} x^{m-1}+...+b_0 are polynomials, then when looking at p(x)/q(x), the long term behavior is the same as that of \frac{a_n x^n}{b_m x^m} because the highest degree term of each polynomial dominates the polynomial when x is large (if you want more rigor, check that e.g. \frac{a_n x^n}{p(x)} goes to 1 as x goes to infinity, so when considering the long term behavior of p(x)/q(x) you can multiply by this without changing it)

Then if n<m, this goes to zero, if n=m this is a_n/b_n and if n>m you get a power function, which isn't really an asymptote unless n=m+1

 

[Mark44]

When finding the asymptote of a rational function, if the degree of the numerator is less than or equal to the denominator you can divide by the highest power in the denominator and then take the limit as x goes to +/- infinity.

Apparently this trick does not work when the numerator is of higher degree and you are finding the slant asymptote, and you must use long division.

When there is a slant asymptote, the rational function is behaving like a straight line with equation y = ax + b, for large or very negative x. A rational function will have a slant asymptote only if the degree of the numerator is one larger than the degree of the denominator. If you carry out the division, you'll get ax + p(x)/q(x), where p(x) is of degree less than that of q(x).

I assume this has something to do with the fact that there isn't really a limit

There won't be a limit in the sense of f(x) (the rational function) approaching some fixed number. As x gets large, f(x) approaches infinity or negative infinity, depending on the signs of the leading terms in the numerator and denominator.

because you aren't approaching a number, but it seems like all the terms with x in their denominator should approach 0 and the x-to-the-something term in the numerator should dominate. I guess the terms with x in their denominator do not approach zero at the same "rate", but why is this not a problem when finding the horizontal asymptote?
I am in my first semester of calculus right now, if that limits the discussion somehow.

If the degree of the denominator is larger than the degree of the numerator, then you can write the denominator as xⁿ(b_n + b_n-1/x + ... + b/x^n-1 + b/xⁿ). If you do the same in the numerator, you'll have xⁿ(a_m/x^{n - m} + a_m-1/x^n-m+1 + ... + a₁/x^n-1 + a₀/xⁿ). You can cancel the xⁿ factors in numerator and denominator, and when you take the limit, the numerator goes to 0 and the denominator goes to infinity (or negative infinity), so the whole fraction goes to zero.

For a more concrete example, consider f(x) = (x² + 2x + 3)/(x⁴ - x² + 5x)

= [x⁴(1/x² + 2/x³ + 3/x⁴]/[x⁴(1 - 1/x² + 5/x³]

The x⁴ factors cancel, and the limit is 0/1 = 0.