## Limits at Infinity

One of our fundamental concepts is the limit of a function at a point. But soon we’ll need to consider what happens as we let the input to a function grow without bound.

So let’s consider a function defined for , where this interval means the set . It really doesn’t matter here what is, just that we’ve got some point where is defined for all larger numbers. We want to come up with a sensible definition for .

When we took a limit at a point we said that if for every there is a so that implies . But this talk of and is all designed to stand in for neighborhoods in a metric space. Picking a defines a neighborhood of the point . All we need is to come up with a notion of a “neighborhood” of .

What we’ll use is a ray just like the one above: . This seems to make sense as the collection of real numbers “near” infinity. So let’s drop it into our definition: the limit of a function at infinity, is if for every there is an so that implies . It’s straightforward to verify from here that this definition of limit satisfies the same laws of limits as the earlier definition.

Finally, we can define neighborhoods of as leftward rays . Then we get a similar definition of the limit of a function at .

One particular limit that’s useful to have as a starting point is . Indeed, given we can set . Then if we see that , establishing the limit.

From here we can handle the limit at infinity of any rational function . Let’s split off the top degree terms from the polynomials and . Divide through top and bottom by to write

Now every term in has degree less than , so each is a multiple of some power of . The laws of limits then tell us that they go to , and the limit of the denominator of is . Thus our limit is the limit of the numerator.

If we have a positive power of as our leading term, which goes up to or down to (depending on the sign of . If , all the powers are negative, and thus the limit is . And if , then all the other powers are negative, and the limit is .

So if the numerator of has the higher degree, we have . If the denominator has higher degree, then . If the degrees are equal, we compare the leading coefficients and find .

There is a bug in numerator of the result limit formula. The first exponent should be $m-n$ and not $n-m$. The further explanation must be also fixed (swapping as appropriate).

Comment by peter | April 18, 2008 |

[...] bigger, trying to fill out the whole ray . And for each one we have a value for the integral: . So we take the limit as approaches infinity: . This will be the value of the integral over the entire [...]

Pingback by Improper Integrals I « The Unapologetic Mathematician | April 18, 2008 |

Thanks peter.

Comment by John Armstrong | April 19, 2008 |

[...] the function is then nondecreasing, and a nondecreasing function bounded above must have a finite limit at infinity. Indeed, the set of values of must be bounded above, and so there is a least upper bound . [...]

Pingback by Convergence Tests for Improper Integrals « The Unapologetic Mathematician | April 21, 2008 |