Epsilon-Delta & Continuity

Let $f$ be defined on an open interval containing $a$, except possibly at $a$ itself. We say $\lim_{x \to a} f(x) = L$ if for every $\varepsilon > 0$, there exists $\delta > 0$ such that

$0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$

We write $f(x) \to L$ as $x \to a$.

Claim: $\lim_{x \to 1} (2x + 1) = 3$.

Proof. Let $\varepsilon > 0$ be given. We need to find $\delta > 0$ such that $0 < |x - 1| < \delta$ implies $|(2x + 1) - 3| < \varepsilon$.

We work backward from the conclusion:

$|(2x + 1) - 3| = |2x - 2| = 2|x - 1|.$

So $|(2x + 1) - 3| < \varepsilon$ is equivalent to $2|x - 1| < \varepsilon$, which is $|x - 1| < \varepsilon/2$.

Choose $\delta = \varepsilon / 2$. Then $0 < |x - 1| < \delta$ implies

$|(2x + 1) - 3| = 2|x - 1| < 2\delta = 2 \cdot \frac{\varepsilon}{2} = \varepsilon.$

$\square$

Claim: $\lim_{x \to 2} x^2 = 4$.

Proof. Let $\varepsilon > 0$ be given. We need $\delta > 0$ such that $0 < |x - 2| < \delta$ implies $|x^2 - 4| < \varepsilon$.

Factor: $|x^2 - 4| = |x - 2| \cdot |x + 2|$. The challenge is that $|x + 2|$ depends on $x$, so we need to bound it. If we restrict $\delta \leq 1$, then $|x - 2| < 1$ means $1 < x < 3$, so $|x + 2| < 5$.

Under this restriction: $|x^2 - 4| = |x - 2| \cdot |x + 2| < |x - 2| \cdot 5$.

Choose $\delta = \min(1, \varepsilon/5)$. Then $0 < |x - 2| < \delta$ implies:

• $\delta \leq 1$, so $|x + 2| < 5$
• $\delta \leq \varepsilon/5$, so $|x - 2| < \varepsilon/5$

Therefore $|x^2 - 4| = |x - 2| \cdot |x + 2| < \frac{\varepsilon}{5} \cdot 5 = \varepsilon$. $\square$

If $\lim_{x \to a} f(x)$ exists, it is unique. That is, if $\lim_{x \to a} f(x) = L_1$ and $\lim_{x \to a} f(x) = L_2$, then $L_1 = L_2$.

Suppose $\lim_{x \to a} f(x) = L_1$ and $\lim_{x \to a} f(x) = L_2$ with $L_1 \neq L_2$. Set $\varepsilon = |L_1 - L_2| / 2 > 0$. By the limit definition, there exist $\delta_1, \delta_2 > 0$ such that:

• $0 < |x - a| < \delta_1 \implies |f(x) - L_1| < \varepsilon$
• $0 < |x - a| < \delta_2 \implies |f(x) - L_2| < \varepsilon$

Let $\delta = \min(\delta_1, \delta_2)$ and pick any $x$ with $0 < |x - a| < \delta$. By the triangle inequality:

$|L_1 - L_2| \leq |f(x) - L_1| + |f(x) - L_2| < \varepsilon + \varepsilon = |L_1 - L_2|.$

This gives $|L_1 - L_2| < |L_1 - L_2|$, a contradiction. Therefore $L_1 = L_2$. $\square$

We say $\lim_{x \to a^-} f(x) = L$ if for every $\varepsilon > 0$, there exists $\delta > 0$ such that

$a - \delta < x < a \implies |f(x) - L| < \varepsilon.$

We say $\lim_{x \to a^+} f(x) = L$ if for every $\varepsilon > 0$, there exists $\delta > 0$ such that

$a < x < a + \delta \implies |f(x) - L| < \varepsilon.$

$\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = L$.

Forward direction. If $\lim_{x \to a} f(x) = L$, then for every $\varepsilon > 0$ there exists $\delta > 0$ such that $0 < |x - a| < \delta$ implies $|f(x) - L| < \varepsilon$. This condition includes both $x \in (a - \delta, a)$ and $x \in (a, a + \delta)$, so both one-sided limits equal $L$.

Reverse direction. Suppose both one-sided limits equal $L$. Given $\varepsilon > 0$, there exist $\delta_1, \delta_2 > 0$ such that:

• $a - \delta_1 < x < a \implies |f(x) - L| < \varepsilon$
• $a < x < a + \delta_2 \implies |f(x) - L| < \varepsilon$

Set $\delta = \min(\delta_1, \delta_2)$. Then $0 < |x - a| < \delta$ means either $a - \delta < x < a$ or $a < x < a + \delta$, and in either case $|f(x) - L| < \varepsilon$. $\square$

We say $\lim_{x \to \infty} f(x) = L$ if for every $\varepsilon > 0$, there exists $M > 0$ such that

$x > M \implies |f(x) - L| < \varepsilon.$

Similarly, $\lim_{x \to -\infty} f(x) = L$ if for every $\varepsilon > 0$, there exists $M > 0$ such that $x < -M$ implies $|f(x) - L| < \varepsilon$.

A function $f$ is continuous at $a$ if:

1. $f(a)$ is defined,
2. $\lim_{x \to a} f(x)$ exists, and
3. $\lim_{x \to a} f(x) = f(a)$.

Equivalently: for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - f(a)| < \varepsilon$.

We say $f$ is continuous on an interval $I$ if $f$ is continuous at every point of $I$.

A function $f$ is continuous at $a$ if and only if for every sequence $(x_n)$ with $x_n \to a$, we have $f(x_n) \to f(a)$.

Forward direction. Assume $f$ is continuous at $a$, and let $(x_n)$ be a sequence with $x_n \to a$. Given $\varepsilon > 0$, by continuity there exists $\delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - f(a)| < \varepsilon$. Since $x_n \to a$, there exists $N$ such that $n \geq N$ implies $|x_n - a| < \delta$. For such $n$, we have $|f(x_n) - f(a)| < \varepsilon$. This proves $f(x_n) \to f(a)$.

Reverse direction (contrapositive). Assume $f$ is not continuous at $a$. Then there exists $\varepsilon_0 > 0$ such that for every $\delta > 0$, there is some $x$ with $|x - a| < \delta$ but $|f(x) - f(a)| \geq \varepsilon_0$. For each $n \in \mathbb{N}$, apply this with $\delta = 1/n$: there exists $x_n$ with $|x_n - a| < 1/n$ but $|f(x_n) - f(a)| \geq \varepsilon_0$. Then $x_n \to a$ (since $|x_n - a| < 1/n \to 0$) but $f(x_n) \not\to f(a)$ (since $|f(x_n) - f(a)| \geq \varepsilon_0$ for all $n$). $\square$

The sequential characterization is the bridge between sequence convergence (Topic 1) and function continuity (this topic). It lets us apply all the sequence-convergence theorems — Monotone Convergence, Squeeze, Bolzano-Weierstrass, the Algebra of Limits — in the function setting. When we need to prove something about a continuous function, we can often reduce it to a statement about convergent sequences, where we already have powerful tools.

If $f$ and $g$ are continuous at $a$, then so are:

1. $f + g$ and $f - g$
2. $f \cdot g$
3. $f / g$ (provided $g(a) \neq 0$)
4. $f \circ g$ (if $g$ is continuous at $a$ and $f$ is continuous at $g(a)$)

We prove this via the sequential characterization. Let $(x_n)$ be any sequence with $x_n \to a$. Since $f$ and $g$ are continuous at $a$, we have $f(x_n) \to f(a)$ and $g(x_n) \to g(a)$.

Sum: By the Algebra of Limits for sequences (Theorem 3 from Sequences, Limits & Convergence), $(f + g)(x_n) = f(x_n) + g(x_n) \to f(a) + g(a) = (f + g)(a)$.

Product: Similarly, $f(x_n) \cdot g(x_n) \to f(a) \cdot g(a)$.

Quotient: If $g(a) \neq 0$, then $f(x_n)/g(x_n) \to f(a)/g(a)$ by the quotient rule for sequences.

Composition: If $g$ is continuous at $a$, then $g(x_n) \to g(a)$. If $f$ is continuous at $g(a)$, then $f(g(x_n)) \to f(g(a))$. This shows $(f \circ g)(x_n) \to (f \circ g)(a)$. $\square$

Every polynomial is continuous on $\mathbb{R}$. Every rational function $p(x)/q(x)$ is continuous on its domain (wherever $q(x) \neq 0$).

In machine learning, this theorem is why compositions of continuous layers produce continuous networks. If each activation function $\sigma_i$ is continuous and each affine map $W_i x + b_i$ is continuous (which it is, as a polynomial), then the full network

$f = \sigma_L \circ W_L \circ \cdots \circ \sigma_1 \circ W_1$

is continuous by iterated application of the composition rule. This is the minimal requirement for the network to have well-defined gradients almost everywhere.

A function $f$ has a removable discontinuity at $a$ if $\lim_{x \to a} f(x)$ exists, but either $f(a)$ is undefined or $f(a) \neq \lim_{x \to a} f(x)$. Redefining $f(a) = \lim_{x \to a} f(x)$ “removes” the discontinuity.

A function $f$ has a jump discontinuity at $a$ if both one-sided limits $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ exist, but $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$.

A function $f$ has an essential discontinuity at $a$ if at least one of the one-sided limits fails to exist (either diverges to infinity or oscillates).

Let $f: [a, b] \to \mathbb{R}$ be continuous. If $y$ is any value between $f(a)$ and $f(b)$, then there exists $c \in (a, b)$ such that $f(c) = y$.

Without loss of generality, assume $f(a) < y < f(b)$ (the case $f(a) > y > f(b)$ follows by applying the argument to $-f$).

Define $S = \{x \in [a, b] : f(x) \leq y\}$. This set is non-empty (it contains $a$, since $f(a) < y \leq y$) and bounded above by $b$. By the completeness of $\mathbb{R}$ (the least upper bound property), $c = \sup S$ exists.

We claim $f(c) = y$. We rule out the other two possibilities:

Case 1: $f(c) < y$. Set $\varepsilon = y - f(c) > 0$. By continuity, there exists $\delta > 0$ such that $|x - c| < \delta$ implies $|f(x) - f(c)| < \varepsilon$, i.e., $f(x) < f(c) + \varepsilon = y$. Since $c < b$ (because $f(b) > y > f(c)$), the interval $(c, c + \delta')$ for $\delta' = \min(\delta, b - c)$ is non-empty, and every $x$ in this interval satisfies $f(x) < y$, hence $x \in S$. But then $c$ is not an upper bound of $S$ — contradiction with $c = \sup S$.

Case 2: $f(c) > y$. Set $\varepsilon = f(c) - y > 0$. By continuity, there exists $\delta > 0$ such that $|x - c| < \delta$ implies $f(x) > f(c) - \varepsilon = y$. So no point in $(c - \delta, c)$ belongs to $S$. But $c = \sup S$ means every interval $(c - \delta, c)$ must contain points of $S$ — contradiction.

Both cases lead to contradictions, so $f(c) = y$. $\square$

If $f$ is continuous on $[a, b]$ and $f(a) \cdot f(b) < 0$ (i.e., $f$ changes sign), then there exists $c \in (a, b)$ with $f(c) = 0$.

The IVT is the theoretical foundation of the bisection method for root-finding. More broadly, it guarantees the existence of decision boundaries: if a continuous classifier assigns a positive score to one data point and a negative score to another, the IVT guarantees that somewhere between them, the classifier outputs exactly zero — a decision boundary exists. This is why continuous classifiers always produce connected decision regions.

If $f: [a, b] \to \mathbb{R}$ is continuous, then $f$ attains its maximum and minimum on $[a, b]$. That is, there exist $c, d \in [a, b]$ such that $f(c) \leq f(x) \leq f(d)$ for all $x \in [a, b]$.

We prove the maximum case; the minimum case follows by applying the argument to $-f$.

Step 1: $f$ is bounded above on $[a, b]$. Suppose not. Then for each $n \in \mathbb{N}$, there exists $x_n \in [a, b]$ with $f(x_n) > n$. The sequence $(x_n)$ is bounded (all terms lie in $[a, b]$), so by the Bolzano-Weierstrass Theorem (from Sequences, Limits & Convergence), there is a convergent subsequence $x_{n_k} \to x^* \in [a, b]$ (the interval is closed, so the limit stays in $[a, b]$). By continuity, $f(x_{n_k}) \to f(x^*)$. But $f(x_{n_k}) > n_k \to \infty$, which contradicts convergence. Therefore $f$ is bounded above.

Step 2: The supremum is attained. Let $M = \sup_{x \in [a,b]} f(x)$, which exists by Step 1. By the definition of supremum, for each $n$ there exists $x_n \in [a, b]$ with $f(x_n) > M - 1/n$. Again by Bolzano-Weierstrass, extract a convergent subsequence $x_{n_k} \to d \in [a, b]$. By continuity, $f(d) = \lim f(x_{n_k})$. Since $M - 1/n_k < f(x_{n_k}) \leq M$ for all $k$, we get $f(d) = M$ by the Squeeze Theorem. $\square$

The EVT is the existence theorem for optimization: it guarantees that $\min_{x \in [a,b]} f(x)$ has a solution when $f$ is continuous and the domain $[a, b]$ is compact (closed and bounded). In machine learning, regularization and weight clipping create compact parameter spaces — for example, constraining $\|\theta\| \leq C$ — precisely to ensure that minimizers exist. Without compactness, a continuous function may approach but never reach its infimum (think of $e^{-x}$ on $(0, \infty)$, which approaches 0 but never reaches it).

A function $f$ is uniformly continuous on a set $D$ if for every $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x, y \in D$:

$|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon.$

A function $f$ is Lipschitz continuous with constant $K \geq 0$ on a set $D$ if

$|f(x) - f(y)| \leq K|x - y| \quad \text{for all } x, y \in D.$

The smallest such $K$ is called the Lipschitz constant of $f$ on $D$.

Lipschitz continuity $\Rightarrow$ uniform continuity $\Rightarrow$ (pointwise) continuity.

Both converses are false.

Lipschitz $\Rightarrow$ uniform continuity. Given $\varepsilon > 0$, choose $\delta = \varepsilon / K$. Then $|x - y| < \delta$ implies $|f(x) - f(y)| \leq K|x - y| < K \cdot \varepsilon/K = \varepsilon$. This $\delta$ depends only on $\varepsilon$ (not on $x$ or $y$), so $f$ is uniformly continuous.

Uniform continuity $\Rightarrow$ continuity. Immediate: uniform continuity at every pair $(x, a)$ with $x$ near $a$ gives pointwise continuity at $a$.

Counterexample for first converse: $f(x) = \sqrt{x}$ on $[0, 1]$ is uniformly continuous (continuous on a compact set — Heine-Cantor, Theorem 5 below) but not Lipschitz: $|f(x) - f(0)| / |x - 0| = 1/\sqrt{x} \to \infty$ as $x \to 0^+$.

Counterexample for second converse: $f(x) = x^2$ on $\mathbb{R}$ is continuous but not uniformly continuous. For any $\delta > 0$, take $x = 1/\delta$ and $y = x + \delta/2$. Then $|x - y| = \delta/2 < \delta$ but $|f(x) - f(y)| = |x^2 - y^2| = |x - y| \cdot |x + y| > (\delta/2)(2/\delta) = 1$. So $\varepsilon = 1$ has no valid $\delta$. $\square$

If $f$ is continuous on a compact set $K$ (closed and bounded in $\mathbb{R}$), then $f$ is uniformly continuous on $K$.

The Heine-Cantor theorem explains why compactness is so important for optimization. On compact domains, continuity automatically gives us uniform continuity — uniform control over function behavior everywhere. This is the machinery behind many existence and approximation theorems in analysis and ML, and we develop it fully in Completeness & Compactness.