You have been using the word continuous since precalculus: a function is continuous if you can draw its graph without lifting your pen. This picture is compelling, but it is not a definition — it is an intuition. And intuitions can fail.
Consider the following:
We now have a rigorous definition of \(\lim_{x\to c}f(x)\) from the last handout. It is time to use it.
A function f: A → ℝ is continuous at a point c ∈ A if, for all ε > 0, there exists δ > 0 such that whenever |x − c| < δ (and x ∈ A): |f(x) − f(c)| < ε. If f is continuous at every point in A, we say f is continuous on A.
Any one of these failing means \(f\) is discontinuous at \(c\).
Reading the picture: Same orange/red band structure as before. The one change: the point \(c\) is now solid on the curve and included in the orange strip. We must have \(f(c)\) land inside the red band — not just nearby points.
Just as functional limits had an \(\varepsilon\)–\(\delta\) version and a sequential version, continuity has multiple equivalent formulations. Each is useful in different situations.
Let \(f: A \to \mathbb{R}\) and \(c \in A\). The function \(f\) is continuous at \(c\) if and only if any one of the following holds:
If \(c\) is a limit point of \(A\), all three are also equivalent to:
| Version | Best used for … |
|---|---|
| \(\varepsilon\)–\(\delta\) | Direct proofs that a specific function IS continuous |
| Sequential (iii) | Proving discontinuity: exhibit one bad sequence \((x_n)\to c\) with \(f(x_n)\not\to f(c)\) |
| (iv) | Inheriting results from the last handout (Algebraic Limit Theorem, etc.) |
Let f: A → ℝ, c ∈ A a limit point. If there exists a sequence (xₙ) ⊆ A with (xₙ) → c but f(xₙ) ↛ f(c), then f is not continuous at c.
Scratch work:
Formal proof:
Note: \(\delta = \varepsilon/3\) works for every \(c\in\mathbb{R}\): it does not depend on \(c\) at all. This will become important when we study uniform continuity.
Discontinuity at integers. Let \(m \in \mathbb{Z}\). We want to show \(h = \lfloor x \rfloor\) is not continuous at \(m\) using the sequential criterion.
Continuity away from integers. Let \(c\notin\mathbb{Z}\), so \(n < c < n+1\) for some \(n\in\mathbb{Z}\). Choose \(\delta = \)
Observation: Here \(\delta\) does not depend on \(\varepsilon\) at all — the same \(\delta\) works for every \(\varepsilon>0\).
Case 1: \(c = 0\).
Case 2: \(c > 0\).
Scratch work:
We do not need to prove continuity from scratch for every function. The Sequential Criterion lets us inherit the Algebraic Limit Theorem:
Assume \(f: A\to\mathbb{R}\) and \(g: A\to\mathbb{R}\) are both continuous at \(c\in A\). Then:
Why? Start from two facts:
By the Algebraic Continuity Theorem (repeatedly applying parts (i)–(iii)):
\(p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n\)
is a sum and product of continuous functions, hence continuous on \(\mathbb{R}\).
By part (iv): rational functions are continuous wherever the denominator is nonzero.
The Algebraic Continuity Theorem handles sums and products. But what about \(\sqrt{3x^2+5}\)? We need to compose two continuous functions. The Algebraic Continuity Theorem does not cover this.
Let \(f: A \to \mathbb{R}\) and \(g: B \to \mathbb{R}\) with \(f(A)\subseteq B\). If \(f\) is continuous at \(c\in A\) and \(g\) is continuous at \(f(c)\in B\), then \(g\circ f\) is continuous at \(c\).
Since \(f\) is continuous at \(c\): \(f(x_n) \to\)
Since \(g\) is continuous at \(f(c)\) and \(f(x_n)\to f(c)\): \(g(f(x_n)) \to\)
That is: \((g\circ f)(x_n) \to g(f(c)) = (g\circ f)(c)\). By the sequential criterion, \(g\circ f\) is continuous at \(c\). \(\square\)
Define a function \(g: \mathbb{R} \to \mathbb{R}\) as follows:
\(g(x) = \begin{cases}1 & x\in\mathbb{Q}\\ 0 & x\notin\mathbb{Q}.\end{cases}\)
Why?