# Finding Asymptotes in Calculus: Rules & Explanation

• shooba
In summary, 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.
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.

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 an/bn and if n>m you get a power function, which isn't really an asymptote unless n=m+1

shooba said:
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).
shooba said:
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.
shooba said:
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(bn + bn-1/x + ... + b/xn-1 + b/xn). If you do the same in the numerator, you'll have xn(am/xn - m + am-1/xn-m+1 + ... + a1/xn-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.

## What is an asymptote in calculus?

An asymptote is a line that a graph approaches but never touches. It is a straight line that represents the limit of a curve as it approaches infinity or a vertical/horizontal line.

## Why is finding asymptotes important in calculus?

Finding asymptotes helps us understand the behavior of a function as it approaches certain values or infinity. It also helps us identify any discontinuities in a function.

## What are the different types of asymptotes?

There are three types of asymptotes: horizontal, vertical, and slant. Horizontal asymptotes occur when the function approaches a constant value as x approaches infinity. Vertical asymptotes occur when the function approaches infinity as x approaches a particular value. Slant asymptotes occur when the function approaches a linear function as x approaches infinity.

## How do you find asymptotes in calculus?

To find asymptotes, you can use the following rules:
1. For horizontal asymptotes, find the limit of the function as x approaches infinity.
2. For vertical asymptotes, find the values of x that make the denominator of the function equal to zero.
3. For slant asymptotes, use long division to divide the numerator by the denominator and the quotient will be the equation of the slant asymptote.

## What are some common mistakes when finding asymptotes in calculus?

Some common mistakes include forgetting to check for vertical asymptotes, not simplifying the function before finding asymptotes, and using the wrong method for finding the type of asymptote. It is important to carefully follow the rules and double check your work to avoid these mistakes.

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