Limits and continuity

Limits and continuity provide the rigorous language for describing how functions behave near a point. They are the foundation on which differentiation and integration are built: every derivative is a limit of a difference quotient, and every integral is a limit of sums. This page covers the ε–δ definition, uniform versus pointwise convergence of function sequences, and the compactness theorems (Bolzano–Weierstrass, Heine–Borel) that turn local control into global control.

Authoritative sources

W. Rudin, Principles of Mathematical Analysis, Ch. 2–4

M. Spivak, Calculus, Ch. 1–3

The ε–δ definition of a limit

Definition. Let $f: D \to \mathbb{R}$ and $a \in \mathbb{R}$ be a limit point of $D$ . We write $\lim_{x \to a} f(x) = L$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x \in D$ with $0 < |x - a| < \delta$ , we have $|f(x) - L| < \varepsilon$ .

Details

Why ε–δ?

Before the 19th century, “limit” meant an intuitive idea: $f(x)$ gets “arbitrarily close” to $L$ as $x$ gets “arbitrarily close” to $a$ . Cauchy (1821) and Weierstrass (1860s) replaced this with a quantifier-based definition that makes the notion checkable.

The logic reads: “No matter how tight a band around $L$ you demand (that is $\varepsilon$ ), I can find a window around $a$ (that is $\delta$ ) so that every input in that window lands in the band.”

Worked example

Prove $\lim_{x \to 2} (3x - 1) = 5$ .

Proof. Let $\varepsilon > 0$ . Choose $\delta = \varepsilon / 3$ . If $0 < |x - 2| < \delta$ , then

$|f(x) - 5| = |3x - 1 - 5| = 3|x - 2| < 3\delta = \varepsilon.$

∎

Continuity

Definition. $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$ . Equivalently: for every $\varepsilon > 0$ there exists $\delta > 0$ such that $|x - a| < \delta \implies |f(x) - f(a)| < \varepsilon$ .

The only difference from the limit definition is that $x = a$ is included and the target is $f(a)$ rather than an abstract $L$ .

Example problem

Problem. Prove that $f(x) = x^2$ is continuous at $x = 3$ .

Solution

We need $|x^2 - 9| < \varepsilon$ whenever $|x - 3| < \delta$ .

Factor: $|x^2 - 9| = |x - 3||x + 3|$ .

If $\delta \leq 1$ , then $|x + 3| \leq |x - 3| + 6 \leq 7$ .

So $|x^2 - 9| \leq 7|x - 3|$ . Choose $\delta = \min(1, \varepsilon/7)$ .

Then $|x - 3| < \delta \implies |x^2 - 9| < 7\delta \leq \varepsilon$ . ∎

Pointwise versus uniform convergence

Definition (pointwise). A sequence $f_n: X \to \mathbb{R}$ converges pointwise to $f$ if for every $x \in X$ and every $\varepsilon > 0$ there exists $N$ (depending on $x$ ) such that $n \geq N \implies |f_n(x) - f(x)| < \varepsilon$ .

Definition (uniform). $f_n \to f$ uniformly on $X$ if for every $\varepsilon > 0$ there exists $N$ (independent of $x$ ) such that $n \geq N \implies |f_n(x) - f(x)| < \varepsilon$ for all $x \in X$ .

Key theorem. If $f_n$ are continuous on $X$ and $f_n \to f$ uniformly, then $f$ is continuous on $X$ . Pointwise convergence does not preserve continuity.

Details

The distinction in one picture

The difference is where $N$ may depend on $x$ :

Pointwise: for each fixed $x$ , the sequence $f_n(x)$ converges. The rate can vary wildly from point to point.
Uniform: the entire graph of $f_n$ gets within $\varepsilon$ of the graph of $f$ simultaneously.

Classic counterexample

On $[0,1]$ , let $f_n(x) = x^n$ . Then $f_n \to f$ pointwise where $f(x) = 0$ for $x \in [0,1)$ and $f(1) = 1$ . The limit is discontinuous at $x = 1$ , so the convergence is not uniform.

To see non-uniformity directly: $\sup_{x \in [0,1]} |x^n - f(x)| = 1$ for every $n$ (take $x = 1$ ), so no $N$ works for $\varepsilon = 1/2$ .

Uniform limit theorem (proof sketch)

Example problem

Problem. Show that $g_n(x) = \frac{x}{n}$ converges uniformly to $0$ on $[0, 1]$ . Does it converge uniformly on $\mathbb{R}$ ?

Solution

On $[0,1]$ : $\sup_{x \in [0,1]} |x/n| = 1/n \to 0$ , so convergence is uniform.

On $\mathbb{R}$ : $\sup_{x \in \mathbb{R}} |x/n| = \infty$ for every $n$ , so convergence is not uniform. It is pointwise (for each fixed $x$ , $x/n \to 0$ ).

This illustrates why the domain matters: uniform convergence is a property of the sequence on a set, not just of the pointwise limit. ∎

Compactness: Bolzano–Weierstrass and Heine–Borel

Definition (Bolzano–Weierstrass). Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

Definition (Heine–Borel). A subset $K \subseteq \mathbb{R}$ is compact if and only if it is closed and bounded.

Theorem (continuous image of compact is compact). If $f: K \to \mathbb{R}$ is continuous and $K$ is compact, then $f(K)$ is compact. In particular, $f$ attains its maximum and minimum on $K$ .

Details

Why compactness matters

Compactness is the property that lets you pass from “every finite subcollection has a property” to “the whole collection has that property.” In $\mathbb{R}$ , it is exactly closed + bounded.

Bolzano–Weierstrass in action. Let $(a_n)$ be bounded. The sequence has a convergent subsequence. Proof sketch: by the nested-interval (bisection) argument, there is a point $a$ such that infinitely many $a_n$ lie in every neighborhood of $a$ ; pick those indices to form the subsequence.

Heine–Borel in action. The set $(0, 1)$ is bounded but not closed, hence not compact. Indeed, the open cover $\{(1/n, 1) : n \geq 2\}$ has no finite subcover.

Extremum theorem. If $f$ is continuous on $[a, b]$ , then $f$ attains a maximum and a minimum. This is the workhorse behind the intermediate value theorem’s “crossing” argument and the existence of solutions in optimization.

Example problem

Problem. Use Bolzano–Weierstrass to show that every bounded sequence in $\mathbb{R}$ has a convergent subsequence. (Give the bisection argument.)

Solution

Let $(a_n)$ be bounded, say $a_n \in [A, B]$ .

Bisect $[A, B]$ into two halves. At least one half contains infinitely many $a_n$ . Call it $I_1$ .
Bisect $I_1$ . At least one half contains infinitely many $a_n$ . Call it $I_2 \subseteq I_1$ .
Continue inductively: $I_{k+1} \subseteq I_k$ , each $I_k$ contains infinitely many $a_n$ , and $|I_k| = (B-A)/2^k \to 0$ .

By the nested-interval property, $\bigcap_k I_k = \{a\}$ for some $a$ . Since each $I_k$ contains infinitely many terms, we can pick $a_{n_k} \in I_k$ with $n_1 < n_2 < \cdots$ . Then $|a_{n_k} - a| \leq |I_k| \to 0$ , so $a_{n_k} \to a$ . ∎

Quiz

Example problem

Problem. Let $f_n: [0,1] \to \mathbb{R}$ be defined by $f_n(x) = x^n(1-x)$ .

(a) Find the pointwise limit $f(x)$ .

(b) Is the convergence uniform? Justify.

Solution

(a) For $x \in [0,1)$ , $x^n \to 0$ , so $f(x) = 0$ . At $x = 1$ , $f_n(1) = 0$ for all $n$ , so $f(1) = 0$ . Thus $f \equiv 0$ .

(b) Compute $\sup_{x \in [0,1]} x^n(1-x)$ . Differentiate: $g(x) = x^n(1-x)$ , $g'(x) = nx^{n-1}(1-x) - x^n = x^{n-1}(n - (n+1)x)$ . Setting $g' = 0$ : $x = n/(n+1)$ . Then

$g(n/(n+1)) = \left(\frac{n}{n+1}\right)^n \cdot \frac{1}{n+1} \approx \frac{1}{e(n+1)} \to 0.$

So $\sup |f_n - f| \to 0$ : the convergence is uniform.

(c) Since $f \equiv 0$ , it is continuous. (Consistent with the uniform limit theorem.) ∎

Limits and continuity §

The ε–δ definition of a limit §

Why ε–δ? §

Worked example §

Continuity §

Pointwise versus uniform convergence §

The distinction in one picture §

Classic counterexample §

Uniform limit theorem (proof sketch) §

Compactness: Bolzano–Weierstrass and Heine–Borel §

Why compactness matters §

Quiz §

Limits and continuity

The ε–δ definition of a limit

Why ε–δ?

Worked example

Continuity

Pointwise versus uniform convergence

The distinction in one picture

Classic counterexample

Uniform limit theorem (proof sketch)

Compactness: Bolzano–Weierstrass and Heine–Borel

Why compactness matters

Quiz