# The Unapologetic Mathematician

## 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. 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| 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. 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, 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 | Analysis, Calculus

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 | Reply

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 [...]

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3. Thanks peter.

Comment by John Armstrong | April 19, 2008 | Reply

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 . [...]

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