(last updated: 10:45:48 AM, October 20, 2020)
Definition 3.1.1 Let \(f\) be a function with domain \(D\subset \mathbb R\), which contains arbitrarily large values. This function \(f\) has a limit as \(x\) approaches (tends to) \(+\infty\) if and only if there exists a real number \(L\) such that for every \(\epsilon>0\), there exists a real number \(M>0\) such that \[|f(x)-L|<\epsilon \quad \text{ for all } \quad x\geq M\] and \(x\in D\). If such an \(L\) exists, we say that \(L\) is the limit of the function \(f\) as \(x\) tends to \(\infty\), or simply, \(L\) is the limit of \(f\) at \(\infty\), or \(L\) is the limiting value at \(\infty\), and we write \(\lim_{x\rightarrow\infty} f(x)=L\), where \(x\) is a dummy variable.
Definition 3.1.2 If \(\lim_{x\rightarrow\infty} f(x)=L\), then \(f\) has a horizontal asymptote at \(+\infty\) and the line \(y=L\) is called a horizontal asymptote for the function \(f\).
Note that we can define limits at \(-\infty\) in the same way as above.
We have already seen “\(\lim\)” notation. Recall that a sequence is just a function whose domain is the natural numbers. So this definition is identical to the definition of convergence fo sequences except that we are now allowing the domain to include other numbers.
It might be tempted to take a sequence of \(x\)-values in the domain \(x_n\) that goes to infinity and consider the sequence \(f(x_n)\). If we can show that \(f(x_n)\) converges to \(L\), then you might think that \(L\) is the limit of function \(f\). Generally, this is ok, but not always. We cannot just pick one sequence! Since the domain of \(f\) can be quite large, say all of \(\mathbb R\), then there will be many different ways to take our sequence of \(x\)-values \(x_n\). The limit of \(f(x_n)\) must not depend on how we choose our sequence of \(x\)-values.
Example: Find the limit of \(f(x)=1-\frac1x\) as \(x\rightarrow\pm\infty\).
We suspect \(\displaystyle\lim_{x\rightarrow\pm\infty} f(x)=1\). Note that \(|f(x)-1|<\epsilon\) requires \(\frac1{|x|}<\epsilon\). So given any \(\epsilon>0\) we choose a cutoff \(x\)-value \(M>\frac1\epsilon\) for \(x\rightarrow+\infty\), or a cutoff of \(-M<-\frac1\epsilon\) for \(x\rightarrow-\infty\) (Noting that I am requiring \(M>0\) here).
Now for the formal argument: Given \(\epsilon>0\), choose any \(M>\frac1\epsilon\). Then for \(x\leq -M\) we have \(|f(x)-1|=\frac1{|x|}\leq\frac1M<\epsilon\) and when \(x\geq M\) we have that \(|f(x)-1|=\frac1{x}\leq\frac1M<\epsilon\). Thus \(f(x)\rightarrow 1\) as \(x\rightarrow\pm\infty\).
Note that a function can have different limits as \(x\rightarrow+\infty\) and \(x\rightarrow-\infty\).
Remark 3.1.4 A function with an unbounded above domain \(D\) has no limiting value at plus infinity; that is, \(f\) has no finite limit at \(+\infty\), if and only if for every real number \(L\) there exists \(\epsilon > 0\) such that for every \(M > 0\) there exists an \(x \geq M\) with \(x \in D\), where \(|f(x) - L | \geq \epsilon\).
This is nearly identical to the statement about a sequence not having a limiting value. The idea is that no matter what \(\epsilon>0\) is given and no matter what cutoff \(x\)-value you choose There will be an \(x\) further out that makes your function value greater than \(\epsilon\) distance away from any proposed limit \(L\).
Example: \(f(x)=\sin(x)\) with domain \(\mathbb R\) has no limiting value as \(x\rightarrow\infty\) since no matter what cutoff \(M\) is chosen there are \(x\geq M\) that have form \(x=\frac\pi2+2\pi\cdot n\) and \(x=\frac{3\pi}2+2\pi\cdot n\) for \(n\in\mathbb N\). So if a limiting value \(L\) is proposed for \(f\), then let \(\epsilon=\frac12\min\{|L-1|,|L+1|\}\) and there will always be \(x\geq M\) (that take one of the forms above) such that \(f(x)=1\) or \(f(x)=-1\) of which at least one of those will be outside of the interval \((L-\epsilon,L+\epsilon)\).
This next result shows that the limit of a function at infinity only exists if every sequence of \(x\)-values in the domain diverging to infinity, when plugged into the function, gives a sequence that converges to the limit.
Theorem 3.1.6 Suppose that \(D\subset \mathbb R\) is an unbounded above domain of the function \(f\); that is, \(D\) contains arbitrarily large values. Then \(\lim_{x\rightarrow a} f (x) = L\) if and only if for every sequence \(x_n\) in \(D\) that diverges to plus infinity, that is, \(\lim_{n\rightarrow\infty} x = \infty\), the sequence \(f(x_n)\) converges to \(L\).
See the textbook for a proof of Theorem 3.1.6.
Theorem 3.1.7 Suppose that the functions \(f\), \(g\), and \(h\) are defined on \(D\subset \mathbb R\), which is unbounded above, with \(\lim_{x\rightarrow\infty} f(x)=A\), \(\lim_{x\rightarrow\infty} g(x)=B\), and \(\lim_{x\rightarrow\infty} h(x)=C\). Then:
The proofs of these are nearly identical to the proofs for the corresponding statements for sequences.
Theorem 3.1.8 If the function f is defined on an unbounded above domain \(D\subset \mathbb R\) and is eventually monotone and eventually bounded, then \(\displaystyle\lim_{x\rightarrow\infty} f(x)\) is finite.
This is essentially the monotone convergenc theorem for functions and is similar to that from Chapter 2.
Definition 3.1.9 Let f be a function with domain \(D \subset\mathbb R\), which contains arbitrarily large values. We say that \(f\) tends to plus infinity as \(x\) tends to \(+\infty\) if and only if for any real \(K > 0\), there exists a real number \(M > 0\) such that \(f (x) > K\) provided that \(x \geq M\) and \(x \in D\). Whenever this is the case, we write \(\displaystyle\lim_{x\rightarrow\infty} f (x) =+\infty\).
Note that this is an “improper” equals sign unless we consider it within the extended real number system. You just need to understand that \(\displaystyle\lim_{x\rightarrow\infty} f (x) =+\infty\) is literally shorthand for the technical statements contained in this definition.
Definition 3.1.10 Let \(f\) be a function with domain \(D\subset\mathbb R\), which contains arbitrarily large negative values. Then \(\lim_{x\rightarrow-\infty} f (x) = L\) if and only if for every \(\epsilon > 0\) there exists a real number \(M> 0\) such that \(|f (x) - L |< \epsilon\) if \(x \leq - M\) and \(x \in D\).
The following lemma is provided in order to give us an elementary way to compare polynomials.
Lemma (not in text) Let \(p(x)\) be a polynomial of degree \(n\). Then \(x^{n+1}\geq p(x)\) eventually. In particular for any \(C\in\mathbb R\), \(x^{n+1}\geq Cx^n\) eventually (for \(x\) sufficiently large and positive). It also follows that if \(m>n\) are natural numbers and \(C\in\mathbb R\), then \(x^m \geq Cx^n\) eventually.
In fact since we have already defined real exponentiation more generally, we have that if \(0<a<b\) are real numbers, then \(x^b \geq Cx^a\) eventually (for large enough positive \(x\)).
Example: Show that \(x^3-5x^2+2x-10 \geq 100x^2+1000\) eventually for large enough positive \(x\).
Solution: Note that this is true if and only if \(x^3+2x \geq 105x^2+1010\) eventually. We know we can make \(x^3\geq 105x^2\) and \(2x \geq 1010\) eventually by the above lemma. Thus the desired result holds.
Now we are ready to introduce rational functions and some results about their asymptotic behavior that you should already be familiar with.
Assume \(n>m\) and consider \(q(x)=\frac{1}{p(x)}\). By part (a) above \(q(x)\overset{x\rightarrow\pm\infty}{\longrightarrow}0\) thus by Theorem 3.1.7(h) we have that \(p(x)=\frac{1}{q(x)}\rightarrow\pm\infty\). The sign will depend on the sign of \(a_n\) only.
More directly, we can divide the top and bottom by \(x^m\) to get \[ f(x)=\frac{a_nx^{n-m}+O(x^{n-m-1})}{b_m+O(\frac1{x^{}})} =\left(\frac{a_n}{b_m}\right) x^{n-m}+O(x^{n-m-1}) \ \overset{x\rightarrow\pm\infty}{\longrightarrow}\ \begin{cases} -\infty & \text{ if } \frac{a_n}{b_m}<0\\ \infty & \text{ if } \frac{a_n}{b_m}>0 \end{cases} \]\(\blacksquare\)
If (c) above, the limit can be either \(\pm\infty\) depending on sign of \(\frac {a_n}{b_m}\) and whether \(n\) is even or odd. There will be several different cases.
The “tail behavior of polynomials” can be thought of as a corollary to Theorem 3.1.13, but these can also be derived directly from the lemma stated above comparing powers of \(x\).
Corollary (not in text) Let \(p(x)\) be a polynomial of degree \(n\) with positive leading coefficient (\(a_n>0\)). If \(n\) is even, then \(\displaystyle\lim_{x\rightarrow\pm\infty}p(x)=+\infty\). If \(n\) is odd, then \(\displaystyle\lim_{x\rightarrow-\infty}p(x)=-\infty\) and \(\displaystyle\lim_{x\rightarrow+\infty}p(x)=+\infty\).
\[ \diamond \S \diamond \]