Limit Laws, Sequential Characterization, and Uniform Continuity
1. Limits and Limit Laws
1.1. Definition and Fundamental Properties
Definition
Given a set \(E \subset {\mathbb R}\) and a function \(f \ : \ E \rightarrow {\mathbb R}\), we say that
\[ \lim_{x \rightarrow p} f(x) = L \]
when for each \(\epsilon > 0\), there is some \(\delta > 0\) such that every \(x \in E\) with \(0 < |x-p| < \delta\) satisfies \(|f(x) - L| < \epsilon\).
Equivalently, \(\lim_{x \rightarrow p} f(x) = L\) if for every open set \(U \subset {\mathbb R}\) containing \(L\), there is some open set \(V \subset {\mathbb R}\) containing \(p\) such that \(f\) maps \(V \cap E \setminus \{p\}\) into \(U\).
Important
There is an important caveat: This definition makes very little practical sense if \(p\) is not a limit point of the set \(E\). Limit points of \(E\) are those points \(q \in {\mathbb R}\) such that every neighborhood of \(q\) contains at least one point of \(E\) other than \(q\) itself. If \(p\) is not a limit point, then the statements
\[ \lim_{x \rightarrow p} f(x) = 1 \text{ and } \lim_{x \rightarrow p} f(x) = 2 \]
are both always true and do not contradict one another!
Theorem (Uniqueness of Limits)
If \(p\) is a limit point of \(E\) and \(f : E \rightarrow {\mathbb R}\), then
\[ \lim_{x \rightarrow p} f(x) = L_1 \text{ and } \lim_{x \rightarrow p} f(x) = L_2 \]
implies that \(L_1 = L_2\). In other words, when \(p\) is a limit point, there is at most one possible value that the limit \(\lim_{x \rightarrow p} f(x)\) can have.
Proof
Suppose \(L_1 \neq L_2\) and let \(\epsilon := |L_1 - L_2|/2 > 0\). We know that there exist \(\delta_1,\delta_2\) such that \(0 < |x-p| < \delta_i\) for \(x \in E\) implies \(|f(x) - L_i| < \epsilon\). Because \(p\) is a limit point, there must be some \(x \in E\) such that \(0 < |x-p| < \min\{\delta_1,\delta_2\}\), so \(|f(x) - L_1| < \epsilon\) and \(|f(x) - L_2| < \epsilon\) as well. By the triangle inequality,
\[{}|L_1 - L_2|{}\]
\[{}\leq |L_1 - f(x)| {}\]
\[{}+ |f(x) - L_2|{}\]
\[{}< 2 \epsilon = |L_1 - L_2|, {}\]
which gives a contradiction.
Another fundamental result is about boundedness:
Theorem
If \(f \ : \ E \rightarrow {\mathbb R}\) and \(\lim_{x \rightarrow p} f(x)\) exists, then there is a neighborhood \(U\) of \(p\) such that \(f\) is bounded on \(E \cap U\). Moreover, if \(\lim_{x \rightarrow p} f(x) \neq 0\) and \(p\) is a limit point of \(E\), then there exists an open set \(U\) containing \(p\) such that \(f(x)\) is bounded away from \(0\) for all \(x \in U \cap E \setminus \{p\}\).
Proof
Suppose the limit equals \(L\) and fix \(\epsilon = 1\); then there is some \(\delta > 0\) such that every \(x \in E\) with \(0 < |x-p| < \delta\) has \(|f(x) - L| < 1\); this means that \(|f(x)| \leq |L|+1\) for all \(x \in E \cap (p-\delta,p+\delta) \setminus \{p\}\). If \(p \not \in E\), then \(E \cap (p-\delta,p+\delta) \setminus \{p\}\) simply equals \(E \cap (p-\delta,p+\delta)\) and taking \(U := (p-\delta,p+\delta)\) satisfies the conditions of the theorem. Otherwise, \(p \in E\). But then \(|f(x)| \leq \max\{|L|+1, |f(p)| \}\) for all \(x \in E \cap (p-\delta,p+\delta)\) because either \(x \neq p\), in which case \(|f(x)| \leq |L|+1\), or \(x = p\), in which case \(|f(x)| \leq |f(p)|\).
For the second claim, we can simply choose \(\delta\) so that \(x \in E\) and \(0 < |x-p| < \delta\) implies \(|f(x) - L| < |L|/2\). This forces \(|f(x)| \geq |L|/2 > 0\) for all \(x \in E \cap (p-\delta,p+\delta) \setminus \{p\}\).
1.2. Limit Laws and Continuity
Theorem (Algebraic Limit Laws)
Suppose that \(E \subset {\mathbb R}\) and that \(f\) and \(g\) are real-valued functions on \(E\) and that
\[ \lim_{x \rightarrow p} f(x) = A \text{ and } \lim_{x \rightarrow p} g(x) = B. \]
Then
- \(\lim_{x \rightarrow p} (f(x) + g(x)) = A + B\).
- For any constant \(c \in {\mathbb R}\), \(\lim_{x \rightarrow p} c f(x) = c A\).
- \(\lim_{x \rightarrow p} f(x) g(x) = A B\).
- If \(B \neq 0\), then \(\lim_{x \rightarrow p} \frac{f(x)}{g(x)} = \frac{A}{B}\).
In particular, the relevant limits are guaranteed to exist.
Proof
Definition
When \(f \ : \ E \rightarrow {\mathbb R}\) and \(p \in E\), we say that \(f\) is continuous at \(p\) when \(\lim_{x \rightarrow p} f(x)\) exists and equals \(f(p)\).
As above, the definition makes sense but is somewhat pointless if \(p\) is not also a limit point of \(E\).
Theorem (Continuity and Limits)
Suppose that \(E, F\) are subsets of the real numbers and that \(g \ : \ E \rightarrow F\) and \(f \ : \ F \rightarrow {\mathbb R}\). If \(\lim_{x \rightarrow p} g(x) = L\), if \(L \in F\), and if \(f\) is continuous at \(L\), then
\[ \lim_{x \rightarrow p} f(g(x)) = f(\lim_{x \rightarrow p} g(x)) = f(L). \]
In other words, the limit passes through functions \(f\) which are continuous at \(L\).
Proof
Fix any \(\epsilon > 0\); there exists some \(\eta\) such that \(y \in F \cap (L-\eta,L+\eta) \setminus \{L\}\) implies \(|f(y) - f(L)| < \epsilon\). Since the inequality \(|f(L) - f(L)| < \epsilon\) is trivially true, any \(y \in F \cap (L-\eta,L+\eta)\) satisfies \(|f(y) - f(L)| < \epsilon\). Now given \(\eta > 0\), there exists \(\delta > 0\) such that \(x \in E \cap (p-\delta,p+\delta) \setminus \{p\}\) implies that \(|g(x) - L| < \eta\). Any such \(x\) must then have \(g(x) \in F \cap (L-\eta,L+\eta)\), and consequently \(|f(g(x))- f(L)| < \epsilon\).
Corollary
If \(g\) is defined and continuous at \(p\) and \(f\) is defined and continuous at \(g(p)\), then \(f(g(x))\) is defined and continuous at \(p\) provided that the domain of \(f\) contains the image via \(g\) of some neighborhood of \(p\).
2. The Sequential Characterization of Limits
For real-valued functions \(f\) defined on sets \(E \subset {\mathbb R}\), it can sometimes be convenient to use the following characterization of limits. It changes the problem of proving existence of a functional limit into one of proving existence of limits of sequences:
Theorem (Sequential Characterization)
Suppose that \(p \in E \subset {\mathbb R}\) and \(f \ : \ E \rightarrow {\mathbb R}\). Then
\[ \lim_{x \rightarrow p} f(x) = L \]
if and only if
\[ \lim_{n \rightarrow \infty} f(x_n) = L \]
for every sequence \(\{x_n\}_{n=1}^\infty \subset E \setminus \{p\}\) such that \(\lim_{n \rightarrow \infty} x_n = p\).
Proof
\([\Rightarrow]\) Suppose that \(\lim_{x \rightarrow p} f(x) = L\). Let \(\{x_n\}_{n=1}^\infty\) be any sequence in \(E \setminus \{p\}\) which converges to \(p\). Then for any \(\epsilon > 0\), there is some \(\delta > 0\) such that \(0 < |y-p| < \delta\) for \(y \in E\) implies \(|f(y) - L| < \epsilon\). Since \(x_n \rightarrow p\), there is some \(N > 0\) such that \(|x_n - p| < \delta\) for all \(n > N\), and since \(x_n\) is known to belong to \(E \setminus \{p\}\), \(0 < |x_n - p| < \delta\) for all \(n > N\). Thus for all \(n > N\), \(|f(x_n) - L| < \epsilon\) by combining the two statements above.
\([\Leftarrow]\) Suppose that \(\lim_{x \rightarrow p} f(x)\) is not equal to \(L\). Then there must exist some \(\epsilon > 0\) such that every \(\delta > 0\) admits some \(x \in E\) with \(0 < |x-p| < \delta\) but \(|f(x) - L| \geq \epsilon\). By taking \(\delta_n := \frac{1}{n}\), there must exist a sequence \(\{x_n\}_{n=1}^\infty\) such that \(x_n \in E\) for each \(n\) and \(0 < |x_n - p| < \frac{1}{n}\) for each \(n\) but \(|f(x_n) - L| \geq \epsilon\). Consequently \(\{x_n\}_{n=1}^\infty\) is itself a sequence in \(E \setminus \{p\}\) which converges to \(p\) but has that \(f(x_n) \not \rightarrow L\) as \(n \rightarrow \infty\). This contradicts the hypothesis that all such sequences \(\{x_n\}_{n=1}^\infty\) must have \(f(x_n) \rightarrow L\) as \(n \rightarrow \infty\).
3. Uniform Continuity
As mentioned previously, uniform continuity is a technical strengthening of the notion of continuity in situations when it is possible to choose a \(\delta\) for a given value of \(\epsilon\) which happens to be simultaneously valid at all points \(x\) in the domain. Generally speaking, this is not possible. A non-example is the function \(x^2\) on \((-\infty,\infty)\). For any \(\epsilon > 0\), the points \(x = n\) and \(y = n + \frac{1}{n}\) have
\[ \left( n + \frac{1}{n} \right)^2 - n^2 = 2 + \frac{1}{n^2} > 2 \]
but \(|(n + \frac{1}{n}) - n| = \frac{1}{n}\). So in particular, no matter the value of \(\delta > 0\), the points \(n + \frac{1}{n}\) and \(n\) are eventually within distance less than \(\delta\) of each other, but the values of the function \(f(x) = x^2\) always differ by at least two at these two points. So for \(f(x) = x^2\) on \((-\infty,\infty)\), it is necessary when proving continuity to choose a \(\delta\) which in some way depends upon the particular point at which continuity is being established.
That said, when the domain is compact, continuity automatically upgrades to uniform continuity:
Theorem (Continuity on Compact Domains Implies Uniform Continuity)
If \(E \subset {\mathbb R}\) is compact, then every continuous function \(f \ : \ E \rightarrow {\mathbb R}\) is uniformly continuous on \(E\).
Proof
Suppose not: there must be some \(\epsilon > 0\) such that for every \(\delta > 0\), there is some pair of points \(x,y \in E\) such that \(|x-y| < \delta\) but \(|f(x) - f(y)| \geq \epsilon\). Let \(\{x_n\}_{n=1}^\infty\) and \(\{y_n\}_{n=1}^\infty\) be two sequences in \(E\) obtaining by fixing \(\delta = \frac{1}{n}\) for \(n=1,2,3,\ldots\), i.e., such that \(|x_n - y_n| < \frac{1}{n}\) but \(|f(x_n) - f(y_n)| \geq \epsilon\). By Bolzano-Weierstrass, one may pass to a subsequence \(\{x_{n_i}\}_{i=1}^\infty\) which converges to some point \(x \in E\) because \(E\) is compact. Moreover, since \(\lim_{n \rightarrow \infty} (x_n - y_n) = 0\), \(\lim_{i \rightarrow \infty} (x_{n_i}-y_{n_i}) = 0\) as well (and the limit is guaranteed to exist). Since \(y_{n_i} = x_{n_i} + (y_{n_i} - x_{n_i})\), \(y_{n_i}\) is the sum of two convergent sequences, \(\{y_{n_i}\}_{i=1}^\infty\) converges to \(x+0 = x\) as \(i \rightarrow \infty\). Since \(f\) is continuous at \(x\),
\[{}\lim_{i \rightarrow \infty} (f(x_{n_i})- f(y_{n_i})){}\]
\[{}= f(x) - f(x) = 0.{}\]
But this contradicts the condition that \(|f(x_{n_i}) - f(y_{n_i})| \geq \epsilon\) for each \(i\), since convergence to zero of \((f(x_{n_i})- f(y_{n_i}))\) mandates that \(|f(x_{n_i}) - f(y_{n_i})| < \epsilon\) for all sufficiently large \(i\).
This will be a technical result which we use only infrequently but to great effect.