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Chapter 4: Functional Limits and Continuity

Limit of Functions

From Sequential to Functional Limits

We have a precise definition of \(\lim_{n\to\infty} a_n = L\) for a sequence. The idea behind \(\lim_{x\to c} f(x) = L\) is analogous: \(f(x)\) gets arbitrarily close to \(L\) whenever \(x\) is chosen close enough (but not equal) to \(c\).

Recall the \(\varepsilon\)–\(N\) definition of limit: for every \(\varepsilon > 0\), there exists \(N \in \mathbb{N}\) such that whenever \(n \ge N\),

\(|a_n-L|<\varepsilon\)

n \(a_n\) L+ε L-ε L
Rewrite the definition of \(a_n \to L\) using \(\varepsilon\)-neighborhoods of \(L\):

For functional limits, the challenge is still an \(\varepsilon\)-neighborhood around \(L\), but the response is a \(\delta\)-neighborhood around \(c\).

The \(\varepsilon\)–\(\delta\) Definition

Definition

Let \(f: A \to \mathbb{R}\), and let \(c\) be a limit point of the domain \(A\). We say \(\lim_{x \to c} f(x) = L\) provided that, for all \(\varepsilon > 0\), there exists \(\delta > 0\) such that whenever

\(0 < |x - c| < \delta\) and \(x \in A\) implies \(|f(x) - L| < \varepsilon\)

Important remarks:

  1. The condition \(0 < |x-c|\) means \(x \neq c\): what happens at \(c\) itself is irrelevant.
  2. The point \(c\) need even be in the domain of \(f\).
  3. We only discuss \(\lim_{x\to c} f(x)\) when \(c\) is a of \(A\) (so \(c\) can be approached).
x f(x) L c L

Reading the picture: The challenge is any \(\varepsilon > 0\) (horizontal band around \(L\)). The response is a \(\delta > 0\) (vertical strip around \(c\)). The definition says: the entire portion of the graph inside the vertical strip (excluding \(x=c\)) must lie inside the horizontal band.

Topological Version

Definition

Let \(c\) be a limit point of the domain of \(f: A \to \mathbb{R}\). We say \(\lim_{x\to c} f(x) = L\) provided that, for every \(\varepsilon\)-neighborhood \(V_\varepsilon(L)\) of \(L\), there exists a \(\delta\)-neighborhood \(V_\delta(c)\) such that for all \(x \in V_\delta(c)\) with \(x \neq c\) and \(x \in A\):

\(f(x) \in V_\varepsilon(L)\)

Examples

Example 1. Prove that \(\lim_{x \to -1}(4x+3) = -1\).

Scratch work:

Formal proof:

x f(x) -1 -1
Example 2. Prove \(\lim_{x\to 2} x^2 = 4\).

Scratch work:

Formal proof:

x f(x) 4 2

Sequential Criterion for Functional Limits

Theorem

Given \(f: A \to \mathbb{R}\) and a limit point \(c\) of \(A\), the following are equivalent:

  1. \(\lim_{x \to c} f(x) = L\)
  2. For all sequences \((x_n) \subseteq A\) with \(x_n \neq c\) and \((x_n) \to c\), we have \(f(x_n) \to L\).

Proof sketch: \((i)\implies (ii)\)

Assume \(\lim_{x\to c} f(x) = L\). Let \((x_n) \to c\) with \(x_n \neq c\).

Use the definition of \(\lim_{x \to c}f(x)=L\) to show that \(f(x_n) \to L\).

Proof sketch: \((ii) \implies (i)\) (by contrapositive)

Assume \(\lim_{x\to c} f(x) \neq L\).

Write what it means to say that \(\lim_{x\to c} f(x) \neq L\).

For each \(n \in \mathbb{N}\), apply the negation above with \(\delta = \frac{1}{n}\) to obtain \(x_n\) with

\(0 < |x_n - c| < \frac{1}{n}\) and \(|f(x_n) - L| \geq \varepsilon_0\)

Explain why \((x_n) \to c\) but \(f(x_n) \not\to L\), and why this contradicts \((ii)\).

Divergence Criterion: Proving Limits Don't Exist

Corollary

Let \(f: A \to \mathbb{R}\) and \(c\) a limit point of \(A\). If there exist sequences \((x_n)\) and \((y_n)\) in \(A\) with \(x_n \neq c\), \(y_n \neq c\), and

\(\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n = c\) but \(\lim_{n \to \infty} f(x_n) \neq \lim_{n \to \infty} f(y_n)\),

then \(\lim_{x\to c} f(x)\) does not exist.

Example 3. Show \(\lim_{x\to 0} \sin(1/x)\) does not exist.

The function oscillates rapidly near \(0\).

x sin(1/x) 1 -1

Take \(x_n = \) and \(y_n = \)

Checklist:

Show that \(x_n \to 0\) and \(y_n \to 0\).

Check \(\sin(1/x_n) \neq \sin(1/y_n)\).

Since \(\lim \sin(1/x_n) = 0 \neq 1 = \lim \sin(1/y_n)\), the limit does not exist. \(\blacksquare\)

The Algebraic Limit Theorem for Functional Limits

Just as we built up arithmetic for sequences, the Sequential Criterion lets us inherit the results immediately:

Corollary (Algebraic Limit Theorem)

Assume \(\lim_{x\to c} f(x) = L\) and \(\lim_{x\to c} g(x) = M\). Then:

  1. \(\lim_{x\to c} kf(x) = \) for all \(k \in \mathbb{R}\)
  2. \(\lim_{x\to c} [f(x) + g(x)] = \)
  3. \(\lim_{x\to c} [f(x)g(x)] = \)
  4. \(\lim_{x\to c} \frac{f(x)}{g(x)} = \) , provided \(M \neq 0\)

Activity

Problem 1. Prove that \(\lim_{x\to 2}(3x+4) = 10\) using the definition of limit.
Problem 2. Decide whether each claim is true or false, and give a brief justification:
  1. Show that if \(\delta > 0\) is a valid choice for a given \(\varepsilon > 0\), then any \(\delta'\) with \(0 < \delta' < \delta\) is also a valid choice for the same \(\varepsilon\).
  2. If \(\lim_{x\to a} f(x) = L\) and \(a \in \text{dom}(f)\), then \(L = f(a)\).