The Unapologetic Mathematician

Mathematics for the interested outsider

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 f(x) defined for x\in\left(a,\infty\right), where this interval means the set \left\{x\in\mathbb{R}|a<x\right\}. It really doesn’t matter here what a is, just that we’ve got some point where f is defined for all larger numbers. We want to come up with a sensible definition for \lim\limits_{x\rightarrow\infty}f(x).

When we took a limit at a point p we said that \lim\limits_{x\rightarrow p}f(x)=L if for every \epsilon>0 there is a \delta>0 so that 0<\left|x-p\right|<p implies \left|f(x)-L\right|<\epsilon. But this talk of \epsilon and \delta is all designed to stand in for neighborhoods in a metric space. Picking a \delta defines a neighborhood of the point p. All we need is to come up with a notion of a “neighborhood” of \infty.

What we’ll use is a ray just like the one above: \left(R,\infty\right). 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, \lim\limits_{x\rightarrow\infty}f(x) is L if for every \epsilon>0 there is an R so that x>R implies \left|f(x)-L\right|<\epsilon. 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 -\infty as leftward rays \left(-\infty,R\right)=\left\{x\in\mathbb{R}|x<R\right\}. Then we get a similar definition of the limit of a function at -\infty.

One particular limit that’s useful to have as a starting point is \lim\limits_{x\rightarrow\infty}\frac{1}{x}=0. Indeed, given \epsilon>0 we can set R=\frac{1}{\epsilon}. Then if x>\frac{1}{\epsilon} we see that \epsilon>\frac{1}{x}>0, establishing the limit.

From here we can handle the limit at infinity of any rational function f(x)=\frac{P(x)}{Q(x)}. Let’s split off the top degree terms from the polynomials P(x)=ax^m+p(x) and Q(x)=bx^n+q(x). Divide through top and bottom by bx^n to write

\displaystyle f(x)=\frac{\frac{a}{b}x^{m-n}+\frac{p(x)}{bx^n}}{1+\frac{q(x)}{bx^n}}

Now every term in q(x) has degree less than n, so each is a multiple of some power of \frac{1}{x}. The laws of limits then tell us that they go to {0}, and the limit of the denominator of f is 1. Thus our limit is the limit of the numerator.

If m>n we have a positive power of x as our leading term, which goes up to \infty or down to -\infty (depending on the sign of \frac{a}{b}. If m<n, all the powers are negative, and thus the limit is {0}. And if m=n, then all the other powers are negative, and the limit is \frac{a}{b}.

So if the numerator of f has the higher degree, we have \lim\limits_{x\rightarrow\infty}f(x)=\pm\infty. If the denominator has higher degree, then \lim\limits_{x\rightarrow\infty}f(x)=0. If the degrees are equal, we compare the leading coefficients and find \lim\limits_{x\rightarrow\infty}f(x)=\frac{a}{b}.

April 17, 2008 - Posted by John Armstrong | Analysis, Calculus | | 4 Comments

4 Comments »

  1. 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

  2. [...] 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

  3. Thanks peter.

    Comment by John Armstrong | April 19, 2008

  4. [...] 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

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