Limits at infinity · Basic Analysis

Section 3.5 Limits at infinity

Note: less than 1 lecture (optional, can safely be omitted unless Section 3.6 or Section 5.5 is also covered)

Subsection 3.5.1 Limits at infinity

Just as with sequences, a continuous variable can also approach infinity.

Definition 3.5.1.

We say \(\infty\) is a cluster point of \(S \subset \R\) if for every \(M \in \R\text{,}\) there exists an \(x \in S\) such that \(x \geq M\text{.}\) Similarly, \(- \infty\) is a cluster point of \(S \subset \R\) if for every \(M \in \R\text{,}\) there exists an \(x \in S\) such that \(x \leq M\text{.}\)

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Let \(f \colon S \to \R\) be a function, where \(\infty\) is a cluster point of \(S\text{.}\) If there exists an \(L \in \R\) with the property that for every \(\epsilon > 0\text{,}\) there is an \(M \in \R\) such that

\begin{equation} \babs{f(x) - L} < \epsilon \end{equation}

whenever \(x \in S\) and \(x \geq M\text{,}\) then we say \(f(x)\) converges to \(L\) as \(x\) goes to \(\infty\text{.}\) We call \(L\) a limit and, if unique, write

\begin{equation} \lim_{x \to \infty} f(x) \coloneqq L . \end{equation}

Alternatively, we write \(f(x) \to L\) as \(x \to \infty\text{.}\)

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Similarly, if \(-\infty\) is a cluster point of \(S\) and there exists an \(L \in \R\) such that for every \(\epsilon > 0\text{,}\) there is an \(M \in \R\) such that

\begin{equation} \babs{f(x) - L} < \epsilon \end{equation}

whenever \(x \in S\) and \(x \leq M\text{,}\) then we say \(f(x)\) converges to \(L\) as \(x\) goes to \(-\infty\text{.}\) Alternatively, we write \(f(x) \to L\) as \(x \to -\infty\text{.}\) We call \(L\) a limit and, if unique, write

\begin{equation} \lim_{x \to -\infty} f(x) \coloneqq L . \end{equation}

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The first thing to do, as usual, is to prove that the limit, if it exists, is unique. We leave it as an exercise for the reader.

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Proposition 3.5.2.

The limit at \(\infty\) or \(-\infty\) as defined above is unique if it exists.

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Example 3.5.3.

Let \(f(x) \coloneqq \frac{1}{\sabs{x}+1}\text{.}\) Then

\begin{equation} \lim_{x\to \infty} f(x) = 0 \qquad \text{and} \qquad \lim_{x\to -\infty} f(x) = 0 . \end{equation}

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Proof: Let \(\epsilon > 0\) be given. Find \(M > 0\) large enough so that \(\frac{1}{M+1} < \epsilon\text{.}\) If \(x \geq M\text{,}\) then \(0 < \frac{1}{\sabs{x}+1} = \frac{1}{x+1} \leq \frac{1}{M+1} < \epsilon\text{.}\) The first limit follows. The proof for \(-\infty\) is left to the reader.

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Example 3.5.4.

Let \(f(x) \coloneqq \sin(\pi x)\text{.}\) Then \(\lim_{x\to\infty} f(x)\) does not exist. To prove this fact, note that if \(x = 2n+\nicefrac{1}{2}\) for some \(n \in \N\text{,}\) then \(f(x)=1\text{,}\) while if \(x = 2n+\nicefrac{3}{2}\text{,}\) then \(f(x)=-1\text{.}\) Both of these values, \(1\) and \(-1\text{,}\) cannot be within a small \(\epsilon\) of a single real number.

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Be careful not to confuse continuous limits with limits of sequences. We could say

\begin{equation} \lim_{n \to \infty} \sin(\pi n) = 0, \qquad \text{but} \qquad \lim_{x \to \infty} \sin(\pi x) \enspace \text{does not exist}. \end{equation}

Of course, the notation is ambiguous: Are we thinking of the sequence \(\bigl\{ \sin (\pi n) \bigr\}_{n=1}^\infty\) or the function \(\sin(\pi x)\) of a real variable? We are simply using the convention that \(n \in \N\text{,}\) while \(x \in \R\text{.}\) When the notation is not clear, it is good to explicitly mention where the variable lives, or what kind of limit you are using. If there is a possibility of confusion, one can write, for example,

\begin{equation} \lim_{\substack{n \to \infty\\n \in \N}} \sin(\pi n) . \end{equation}

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There is a connection between continuous limits and limits of sequences, but we must take all sequences going to infinity, just as before in Lemma 3.1.7.

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Lemma 3.5.5.

Suppose \(f \colon S \to \R\) is a function, \(\infty\) is a cluster point of \(S \subset \R\text{,}\) and \(L \in \R\text{.}\) Then

\begin{equation} \lim_{x\to\infty} f(x) = L \qquad \text{if and only if} \qquad \lim_{n\to\infty} f(x_n) = L \end{equation}

for all sequences \(\{ x_n \}_{n=1}^\infty\) in \(S\) such that \(\lim\limits_{n\to\infty} x_n = \infty\text{.}\)

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The lemma also holds for the limit as \(x \to -\infty\text{.}\) Its proof is almost identical and is left as an exercise.

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Proof.

First suppose \(f(x) \to L\) as \(x \to \infty\text{.}\) Given an \(\epsilon > 0\text{,}\) there exists an \(M\) such that for all \(x \in S\) where \(x \geq M\text{,}\) we have \(\babs{f(x)-L} < \epsilon\text{.}\) Let \(\{ x_n \}_{n=1}^\infty\) be a sequence in \(S\) such that \(\lim_{n\to\infty} x_n = \infty\text{.}\) Then there exists an \(N\) such that \(x_n \geq M\) for all \(n \geq N\text{.}\) Thus \(\babs{f(x_n)-L} < \epsilon\text{.}\)

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We prove the converse by contrapositive. Suppose \(f(x)\) does not go to \(L\) as \(x \to \infty\text{.}\) This means that there exists an \(\epsilon > 0\) such that for every \(n \in \N\text{,}\) there exists an \(x \in S\text{,}\) \(x \geq n\text{,}\) let us call it \(x_n\text{,}\) such that \(\babs{f(x_n)-L} \geq \epsilon\text{.}\) Consider the sequence \(\{ x_n \}_{n=1}^\infty\text{.}\) Clearly \(\bigl\{ f(x_n) \bigr\}_{n=1}^\infty\) does not converge to \(L\text{.}\) It remains to note that \(\lim_{n\to\infty} x_n = \infty\) because \(x_n \geq n\) for all \(n\text{.}\)

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Using the lemma, we again translate results about sequential limits into results about continuous limits as \(x\) goes to infinity. That is, we have almost immediate analogues of the corollaries in Subsection 3.1.3. We simply allow the cluster point \(c\) to be either \(\infty\) or \(-\infty\text{,}\) in addition to a real number. We leave it to the student to verify these statements.

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Subsection 3.5.2 Infinite limit

Just as for sequences, it is often convenient to distinguish certain divergent sequences, and talk about limits being infinite almost as if the limits existed.

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Definition 3.5.6.

Let \(f \colon S \to \R\) be a function and suppose \(S\) has \(\infty\) as a cluster point. We say \(f(x)\) diverges to infinity as \(x\) goes to \(\infty\) if for every \(N \in \R\) there exists an \(M \in \R\) such that

\begin{equation} f(x) > N \end{equation}

whenever \(x \in S\) and \(x \geq M\text{.}\) We write

\begin{equation} \lim_{x \to \infty} f(x) \coloneqq \infty , \end{equation}

or we say that \(f(x) \to \infty\) as \(x \to \infty\text{.}\)

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A similar definition can be made for limits as \(x \to -\infty\) or as \(x \to c\) for a finite \(c\text{.}\) Also, similar definitions can be made for limits being \(-\infty\text{.}\) Stating these definitions is left as an exercise. Note that sometimes the slightly incorrect (in this context) wording converges to infinity is used. We can again use sequential limits, and an analogue of Lemma 3.1.7 is left as an exercise.

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Example 3.5.7.

Let us show that \(\lim\limits_{x \to \infty} \frac{1+x^2}{1+x} = \infty\text{.}\)

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Proof: For \(x \geq 1\text{,}\) we have

\begin{equation} \frac{1+x^2}{1+x} \geq \frac{x^2}{x+x} = \frac{x}{2} . \end{equation}

Given \(N \in \R\text{,}\) take \(M = \max \{ 2N+1 , 1 \}\text{.}\) If \(x \geq M\text{,}\) then \(x \geq 1\) and \(\nicefrac{x}{2} > N\text{.}\) So

\begin{equation} \frac{1+x^2}{1+x} \geq \frac{x}{2} > N . \end{equation}

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Subsection 3.5.3 Compositions

Finally, just as for limits at finite numbers, limits play nice with compositions given a continuity assumption.

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Proposition 3.5.8.

Suppose \(f \colon A \to B\text{,}\) \(g \colon B \to \R\text{,}\) \(A, B \subset \R\text{,}\) \(a \in \R \cup \{ -\infty, \infty\}\) is a cluster point of \(A\text{,}\) and \(b \in \R \cup \{ -\infty, \infty\}\) is a cluster point of \(B\text{.}\) Suppose

\begin{equation} \lim_{x \to a} f(x) = b\qquad \text{and} \qquad \lim_{y \to b} g(y) = c \end{equation}

for some \(c \in \R \cup \{ -\infty, \infty \}\text{.}\) If \(b \in B\text{,}\) then suppose \(g(b) = c\text{.}\) Then

\begin{equation} \lim_{x \to a} g\bigl(f(x)\bigr) = c . \end{equation}

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The proof is straightforward, and left as an exercise. We already know the proposition when \(a, b, c \in \R\) (see Exercises 3.1.9 and 3.1.14). Again, the requirement that \(g\) is continuous at \(b\text{,}\) if \(b \in B\text{,}\) is necessary.

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Example 3.5.9.

Let \(h(x) \coloneqq e^{-x^2+x}\text{.}\) Then

\begin{equation} \lim_{x\to \infty} h(x) = 0 . \end{equation}

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Proof: The claim follows once we know

\begin{equation} \lim_{x\to \infty} (-x^2+x) = -\infty \end{equation}

and

\begin{equation} \lim_{y\to -\infty} e^y = 0 , \end{equation}

which is usually proved when the exponential function is defined.

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Exercises 3.5.4 Exercises

3.5.1.

Prove Proposition 3.5.2.

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3.5.2.

Let \(f \colon [1,\infty) \to \R\) be a function. Define \(g \colon (0,1] \to \R\) via \(g(x) \coloneqq f(\nicefrac{1}{x})\text{.}\) Using the definitions of limits directly, show that \(\lim_{x\to 0^+} g(x)\) exists if and only if \(\lim_{x\to \infty} f(x)\) exists, in which case they are equal.

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3.5.3.

Prove Proposition 3.5.8.

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3.5.4.

Let us justify terminology. Let \(f \colon \R \to \R\) be a function such that \(\lim_{x \to \infty} f(x) = \infty\) (diverges to infinity). Show that \(f(x)\) diverges (i.e. does not converge) as \(x \to \infty\text{.}\)

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3.5.5.

Come up with the definitions for limits of \(f(x)\) going to \(-\infty\) as \(x \to \infty\text{,}\) \(x \to -\infty\text{,}\) and as \(x \to c\) for a finite \(c \in \R\text{.}\) Then state the definitions for limits of \(f(x)\) going to \(\infty\) as \(x \to -\infty\text{,}\) and as \(x \to c\) for a finite \(c \in \R\text{.}\)

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3.5.6.

Suppose \(P(x) \coloneqq x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\) is a monic polynomial of degree \(n \geq 1\) (monic means that the coefficient of \(x^n\) is 1).

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Show that if \(n\) is even, then \(\lim\limits_{x\to\infty} P(x) = \lim\limits_{x\to-\infty} P(x) = \infty\text{.}\)
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Show that if \(n\) is odd, then \(\lim\limits_{x\to\infty} P(x) = \infty\) and \(\lim\limits_{x\to-\infty} P(x) = -\infty\) (see previous exercise).
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3.5.7.

Let \(\{ x_n \}_{n=1}^\infty\) be a sequence. Consider \(S \coloneqq \N \subset \R\text{,}\) and \(f \colon S \to \R\) defined by \(f(n) \coloneqq x_n\text{.}\) Show that the two notions of limit,

\begin{equation} \lim_{n\to\infty} x_n \qquad \text{and} \qquad \lim_{x\to\infty} f(x) \end{equation}

are equivalent. That is, show that if one exists so does the other one, and in this case they are equal.

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3.5.8.

Extend Lemma 3.5.5 as follows. Suppose \(S \subset \R\) has a cluster point \(c \in \R\text{,}\) \(c = \infty\text{,}\) or \(c = -\infty\text{.}\) Let \(f \colon S \to \R\) be a function and suppose \(L = \infty\) or \(L = -\infty\text{.}\) Show that

\begin{equation} \begin{aligned} & \lim_{x\to c} f(x) = L \\ & \qquad \text{if and only if} \\ & \lim_{n\to\infty} f(x_n) = L \enspace \text{for all sequences } \{ x_n \}_{n=1}^\infty \text{ such that } \lim_{n\to\infty} x_n = c \text{ and } x_n \in S \setminus \{c\} \text{ for all } n . \end{aligned} \end{equation}

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3.5.9.

Suppose \(f \colon \R \to \R\) is a 2-periodic function, that is \(f(x + 2) = f(x)\) for all \(x\text{.}\) Define \(g \colon \R \to \R\) by \(g(0) \coloneqq f(0)\) and for \(x \neq 0\) by

\begin{equation} g(x) \coloneqq f\left(\frac{\sqrt{x^2+1}-1}{x}\right) \end{equation}

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Find the function \(\varphi \colon (-1,1) \to \R\) such that \(g\bigl(\varphi(t)\bigr) = f(t)\text{,}\) that is, \(\varphi^{-1}(x) = \frac{\sqrt{x^2+1}-1}{x}\) for \(x \neq 0\text{.}\)
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Show that \(f\) is continuous if and only if \(g\) is continuous and

\begin{equation} \lim_{x \to \infty} g(x) = \lim_{x \to -\infty} g(x) = f(1) = f(-1) . \end{equation}

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