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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

As for sequences, a continuous variable can also approach infinity. Let us make this notion precise.

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 \(x \leq M\text{.}\)

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

\begin{equation*} \abs{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 \(L\) the limit and write

\begin{equation*} \lim_{x \to \infty} f(x) := L . \end{equation*}

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

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*} \abs{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 \(-\infty\text{.}\) We call \(L\) the limit and write

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

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

We cheated a little bit again and said the limit. We leave it as an exercise for the reader to prove the following proposition.

Example 3.5.3.

Let \(f(x) := \frac{1}{\abs{x}+1}\text{.}\) Then

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

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 \(\frac{1}{x+1} \leq \frac{1}{M+1} < \epsilon\text{.}\) Since \(\frac{1}{\abs{x}+1} > 0\) for all \(x\) the first limit is proved. The proof for \(-\infty\) is left to the reader.

Example 3.5.4.

Let \(f(x) := \sin(\pi x)\text{.}\) 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 \(f(x)=-1\text{.}\) So they cannot both be within a small \(\epsilon\) of a single real number.

We must 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 \(\{ \sin (\pi n) \}_{n=1}^\infty\) 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 are you using. If there is possibility of confusion, one can write, for example,

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

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

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

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 \(x \geq M\text{,}\) we have \(\abs{f(x)-L} < \epsilon\text{.}\) Let \(\{ x_n \}\) be a sequence in \(S\) such that \(\lim \, x_n = \infty\text{.}\) Then there exists an \(N\) such that for all \(n \geq N\text{,}\) we have \(x_n \geq M\text{.}\) And thus \(\abs{f(x_n)-L} < \epsilon\text{.}\)

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 \(\epsilon > 0\text{,}\) 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 \(\abs{f(x_n)-L} \geq \epsilon\text{.}\) Consider the sequence \(\{ x_n \}\text{.}\) Clearly \(\{ f(x_n) \}\) does not converge to \(L\text{.}\) It remains to note that \(\lim\, x_n = \infty\text{,}\) because \(x_n \geq n\) for all \(n\text{.}\)

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.

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.

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 \(\infty\text{,}\) 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) := \infty , \end{equation*}

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

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 converges to infinity is used. We can again use sequential limits, and an analogue of Lemma 3.1.7 is left as an exercise.

Example 3.5.7.

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

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*}

Subsection 3.5.3 Compositions

Finally, just as for limits at finite numbers we can compose functions easily.

The proof is straightforward, and left as an exercise. We already know the proposition when \(a, b, c \in \R\text{,}\) 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.

Example 3.5.9.

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

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

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.

Subsection 3.5.4 Exercises

Exercise 3.5.2.

Let \(f \colon [1,\infty) \to \R\) be a function. Define \(g \colon (0,1] \to \R\) via \(g(x) := f(\nicefrac{1}{x})\text{.}\) 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.

Exercise 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 \(x \to \infty\text{.}\)

Exercise 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 \(c \in \R\text{.}\)

Exercise 3.5.6.

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

  1. Show that if \(n\) is even, then \(\lim_{x\to\infty} P(x) = \lim_{x\to-\infty} P(x) = \infty\text{.}\)

  2. Show that if \(n\) is odd, then \(\lim_{x\to\infty} P(x) = \infty\) and \(\lim_{x\to-\infty} P(x) = -\infty\) (see previous exercise).

Exercise 3.5.7.

Let \(\{ x_n \}\) be a sequence. Consider \(S := \N \subset \R\text{,}\) and \(f \colon S \to \R\) defined by \(f(n) := x_n\text{.}\) 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.

Exercise 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*} \lim_{x\to c} f(x) = L \qquad \text{if and only if} \qquad \lim_{n\to\infty} f(x_n) = L \enspace \text{for all sequences } \{ x_n \} \text{ such that } \lim\, x_n = c . \end{equation*}

Exercise 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

\begin{equation*} g(x) := f\left(\frac{\sqrt{x^2+1}-1}{x}\right) \end{equation*}
  1. Find the function \(\varphi \colon (-1,1) \to \R\) such that \(g\bigl(\varphi(t)\bigr) = f(t)\text{,}\) that is \(\varphi^{-1}(x) = \frac{\sqrt{x^2+1}-1}{x}\text{.}\)

  2. 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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