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
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) \coloneqq 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
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*}
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 \(0 < \frac{1}{\abs{x}+1} = \frac{1}{x+1} \leq \frac{1}{M+1} < \epsilon\text{.}\) The first limit follows. The proof for \(-\infty\) is left to the reader.
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 \(f(x)=-1\text{.}\) So they cannot both be within a small \(\epsilon\) of a single real number.
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 are you using. If there is possibility of confusion, one can write, for example,
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 \(\{ x_n \}_{n=1}^\infty\) be a sequence in \(S\) such that \(\lim_{n\to\infty} x_n = \infty\text{.}\) Then there exists an \(N\) such that 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 \(\{ x_n \}_{n=1}^\infty\text{.}\) Clearly \(\bigl\{ f(x_n) \bigr\}_{n=1}^\infty\) does not converge to \(L\text{.}\) It remains to note that \(\lim_{n\to\infty} x_n = \infty\text{,}\) 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.
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.
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
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.
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
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.
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.
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 \(x \to \infty\text{.}\)
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 \(c \in \R\text{.}\)
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).
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,
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 .
\end{aligned}
\end{equation*}
Find the function \(\varphi \colon (-1,1) \to \R\) such that \(g\bigl(\varphi(t)\bigr) = f(t)\text{,}\) that is \(\varphi^{-1}(x) =
\frac{\sqrt{x^2+1}-1}{x}\text{.}\)
