How do you find horizontal asymptotes in calculus?

Question

amistre64 · Answer

Hmmm.... i dont quite know what you are asking.  Asymptotes tend to be stationary, and calulus deals with movement.  "HA"s are found from the original function and tend to be limits of the function.

anonymous · Answer

Say for the question 2x/(x^2+9), what is the "calculus method" of getting the H.A.? There is something in my notes about dividing by the largest power and dividing the numerator by the denominator.

amistre64 · Answer

Dividing by the largest power is the algebra way.

In calculus, you can derive another equation from the original to tell you how the original is affected by change; but that does not effect the "HA"s of the original equation in any way, nor does it help you to find them.

For example, the HA of 2/x is y=0.  But the derivative of (2/x) is -2/x^2 which doesnt help in finding the HA.

anonymous · Answer

Ok thanks!

I have this example too that didn't make much sense:

x+1    (divide by x)
lim              ----
x->infinite x-2   (divide by x)

1+1/x  -->0
= lim -----
         1 - 2/x -->0

H.A. = 1

anonymous · Answer

Wait.. I know what its getting at but still.....

amistre64 · Answer

There are 3 simple rules for finding HAs, but they are not calculus. They are just generalizations after doing quite a few algebra modifications.

amistre64 · Answer

If the highest power of x is in the bottom; then the HA =0

If the highest power of x is the same in the top and bottom, the HA=the leading coeficients:  example, 5x^2/7x^2.  HA=5/7

If the highest power of x is found in the top; then the asymptote is a line or curve and not horizontal

anonymous · Answer

Ah that helps me out alot! Thank you for helping me :D