Continuity

4.1 Limits of Functions

Let $f : D \to \mathbb{R}$ where $D \subseteq \mathbb{R}$ . We say $\lim_{x \to a} f(x) = L$ if for Every $\varepsilon > 0$ There exists $\delta > 0$ such that

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

4.2 Continuity

Definition. $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$ . In epsilon-delta form: For every $\varepsilon > 0$ There exists $\delta > 0$ such that

$|x - a| \lt \delta \implies |f(x) - f(a)| \lt \varepsilon$

Remark. A function is continuous on a set $E$ if it is continuous at every point of $E$ . A function is globally continuous (or “continuous”) if it is continuous on its entire domain.

Definition. $f$ is discontinuous at $a$ if it is not continuous at $a$ . Discontinuities are Classified as:

Removable: $\lim_{x \to a} f(x)$ exists but does not equal $f(a)$ (or $f(a)$ is undefined).
Jump: $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ both exist but are unequal.
Essential (or infinite/oscillatory): At least one one-sided limit does not exist.

Proposition 4.3. Polynomials are continuous on $\mathbb{R}$ . Rational functions $p(x)/q(x)$ are Continuous wherever $q(x) \neq 0$ . The functions $\sin x$ , $\cos x$ , $e^x$ , $\ln x$ are continuous On their domains.

Theorem 4.1 (Algebra of Continuous Functions). If $f$ and $g$ are continuous at $a$ Then $f+g$ $f-g$ , $fg$ And (where defined) $f/g$ are continuous at $a$ .

Theorem 4.2. Compositions of continuous functions are continuous: if $f$ is continuous at $a$ and $g$ is continuous at $f(a)$ Then $g \circ f$ is continuous at $a$ .

4.2a Sequential Characterization of Limits and Continuity

The epsilon-delta definitions can be reformulated in terms of sequences, which is often more Convenient for …/1-number-and-algebra/3_proof-and-logics.

Proposition 4.2a (Sequential Criterion for Limits). $\lim_{x \to c} f(x) = L$ if and only if For every sequence $(x_n)$ with $x_n \to c$ and $x_n \neq c$ for all $n$ We have $f(x_n) \to L$ .

Proof. ( $\Rightarrow$ ) Let $\varepsilon > 0$ . Choose $\delta > 0$ from the $\varepsilon$ - $\delta$ definition. Since $x_n \to c$ There exists $N$ with $|x_n - c| \lt \delta$ for $n \geq N$ . Then $|f(x_n) - L| \lt \varepsilon$ for $n \geq N$ .

( $\Leftarrow$ ) Suppose the $\varepsilon$ - $\delta$ condition fails. Then there exists $\varepsilon > 0$ such That for every $n \in \mathbb{N}$ There exists $x_n$ with $0 \lt |x_n - c| \lt 1/n$ but $|f(x_n) - L| \geq \varepsilon$ . Then $x_n \to c$ but $f(x_n) \not\to L$ Contradicting the hypothesis. $\blacksquare$

Corollary 4.2b. $f$ is continuous at $c$ if and only if for every sequence $(x_n)$ with $x_n \to c$ We have $f(x_n) \to f(c)$ .

This is especially useful for proving that a function is not continuous: find one sequence Converging to $c$ whose image does not converge to $f(c)$ .

4.3 Intermediate Value Theorem

Theorem 4.3 (IVT). If $f : [a,b] \to \mathbb{R}$ is continuous and $f(a) \lt y \lt f(b)$ (or $f(b) \lt y \lt f(a)$ ), then there exists $c \in (a,b)$ such that $f(c) = y$ .

Proof. Assume $f(a) \lt y \lt f(b)$ . Let $S = \{x \in [a,b] : f(x) \lt y\}$ . Since $a \in S$ $S$ is non-empty and bounded above by $b$ . Let $c = \sup(S)$ . We show $f(c) = y$ .

If $f(c) \lt y$ Then by continuity at $c$ There exists $\delta > 0$ such that $f(x) \lt y$ for $x \in (c - \delta, c + \delta)$ . But then $c + \delta/2 \in S$ Contradicting that $c = \sup(S)$ .

If $f(c) > y$ Then by continuity, there exists $\delta > 0$ such that $f(x) > y$ for $x \in (c - \delta, c + \delta)$ . But then $c - \delta/2$ is an upper bound for $S$ Contradicting That $c = \sup(S)$ .

Therefore $f(c) = y$ . $\blacksquare$

Alternative …/1-number-and-algebra/3_proof-and-logic (bisection). Set $a_0 = a$ , $b_0 = b$ . Given $[a_n, b_n]$ with $f(a_n) \lt y \lt f(b_n)$ Let $m_n = (a_n + b_n)/2$ . If $f(m_n) \geq y$ Set $a_{n+1} = a_n$ , $b_{n+1} = m_n$ . If $f(m_n) \lt y$ Set $a_{n+1} = m_n$ , $b_{n+1} = b_n$ . Either way, $f(a_n) \lt y \leq f(b_n)$ and $b_n - a_n = (b-a)/2^n \to 0$ . By the nested interval property, $a_n \to c$ and $b_n \to c$ . By continuity, $f(c) = \lim f(a_n) \leq y$ And $f(c) = \lim f(b_n) \geq y$ So $f(c) = y$ . $\blacksquare$

4.4 Extreme Value Theorem

Theorem 4.4 (EVT). If $f : [a,b] \to \mathbb{R}$ is continuous, then $f$ attains its maximum and Minimum on $[a,b]$ : there exist $c_1, c_2 \in [a,b]$ such that $f(c_1) \leq f(x) \leq f(c_2)$ for all $x \in [a,b]$ .

Proof. We first show $f$ is bounded. Suppose not; then for each $n \in \mathbb{N}$ There exists $x_n \in [a,b]$ with $|f(x_n)| > n$ . By Bolzano-Weierstrass, $(x_n)$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$ . By continuity, $f(x_{n_k}) \to f(c)$ So $(f(x_{n_k}))$ is bounded. But $|f(x_{n_k})| > n_k \to \infty$ A contradiction.

Now we show $f$ attains its supremum. Let $M = \sup\{f(x) : x \in [a,b]\}$ . For each $n$ Choose $x_n \in [a,b]$ with $f(x_n) > M - 1/n$ . By Bolzano-Weierstrass, $(x_n)$ has a subsequence $x_{n_k} \to c \in [a,b]$ . By continuity, $f(c) = \lim f(x_{n_k})$ . Since $M - 1/n_k \lt f(x_{n_k}) \leq M$ for all $k$ The squeeze theorem gives $f(c) = M$ . The argument for the infimum is similar (consider $-f$ ). $\blacksquare$

4.5 Uniform Continuity

Definition. $f$ is uniformly continuous on $D$ if for every $\varepsilon > 0$ There exists $\delta > 0$ such that for all $x, y \in D$ :

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

The key distinction: for ordinary continuity, $\delta$ may depend on both $\varepsilon$ and the point $a$ ; for uniform continuity, $\delta$ depends only on $\varepsilon$ .

4.6 The Heine-Cantor Theorem

Theorem 4.5 (Heine-Cantor). If $f : [a,b] \to \mathbb{R}$ is continuous on the closed, bounded Interval $[a,b]$ Then $f$ is uniformly continuous on $[a,b]$ .

Proof. Suppose $f$ is continuous on $[a,b]$ but not uniformly continuous. Then there exists $\varepsilon > 0$ such that for every $n \in \mathbb{N}$ There exist $x_n, y_n \in [a,b]$ with $|x_n - y_n| \lt 1/n$ but $|f(x_n) - f(y_n)| \geq \varepsilon$ .

By the Bolzano-Weierstrass theorem, $(x_n)$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$ . Since $|x_{n_k} - y_{n_k}| \lt 1/n_k \to 0$ We have $y_{n_k} \to c$ as well.

By continuity of $f$ at $c$ : there exists $\delta > 0$ such that $|x - c| \lt \delta$ implies $|f(x) - f(c)| \lt \varepsilon/2$ . For $k$ sufficiently large, $|x_{n_k} - c| \lt \delta$ and $|y_{n_k} - c| \lt \delta$ So:

$|f(x_{n_k}) - f(y_{n_k})| \leq |f(x_{n_k}) - f(c)| + |f(y_{n_k}) - f(c)| \lt \varepsilon/2 + \varepsilon/2 = \varepsilon$

Contradicting $|f(x_{n_k}) - f(y_{n_k})| \geq \varepsilon$ . $\blacksquare$

4.7 Worked Examples

Problem. Prove that $f(x) = \sqrt{x}$ is uniformly continuous on $[0, \infty)$ .

Solution. For $x, y \geq 0$ : $|\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{\sqrt{x} + \sqrt{y}} \leq |x - y|^{1/2}$ .

Given $\varepsilon > 0$ Choose $\delta = \varepsilon^2$ . Then $|x - y| \lt \delta$ implies $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x-y|} \lt \sqrt{\delta} = \varepsilon$ . Since $\delta$ depends Only on $\varepsilon$ The continuity is uniform. $\blacksquare$

Worked Example: $\varepsilon$-$\delta$ .../1-number-and-algebra/3_proof-and-logic that $f(x) = 3x - 1$ is continuous at $x = 2$

Solution. We have $f(2) = 5$ . Let $\varepsilon > 0$ . We need to find $\delta > 0$ such that $|x - 2| \lt \delta$ implies $|f(x) - 5| \lt \varepsilon$ .

Compute: $|f(x) - 5| = |(3x - 1) - 5| = |3x - 6| = 3|x - 2|$ .

Choose $\delta = \varepsilon/3$ . Then $|x - 2| \lt \delta$ implies $|f(x) - 5| = 3|x - 2| \lt 3 \cdot \varepsilon/3 = \varepsilon$ . $\blacksquare$

Worked Example: $\varepsilon$-$\delta$ .../1-number-and-algebra/3_proof-and-logic that $f(x) = x^2$ is continuous at $x = 3$

Solution. We have $f(3) = 9$ . Let $\varepsilon > 0$ . Compute:

$|f(x) - 9| = |x^2 - 9| = |x + 3| \cdot |x - 3|$

Restrict to $\delta \leq 1$ So $|x - 3| \lt 1$ means $2 \lt x \lt 4$ Giving $|x + 3| \lt 7$ .

Choose $\delta = \min(1, \varepsilon/7)$ . Then $|x - 3| \lt \delta$ implies:

$|x^2 - 9| = |x + 3| \cdot |x - 3| \lt 7 \cdot \frac{\varepsilon}{7} = \varepsilon$

$\blacksquare$

Worked Example: Show $f(x) = 1/x$ is NOT uniformly continuous on $(0, 1)$

Solution. We show the negation of uniform continuity. Take $\varepsilon = 1$ . For any $\delta > 0$ Choose $n \in \mathbb{N}$ with $1/n \lt \delta$ . Set $x = 1/n$ and $y = 1/(2n)$ . Then $|x - y| = 1/(2n) \lt 1/n \lt \delta$ But:

$|f(x) - f(y)| = \left|\frac{1}{1/n} - \frac{1}{1/(2n)}\right| = |n - 2n| = n \geq 1 = \varepsilon$

So no single $\delta$ works for all $x, y \in (0,1)$ . $\blacksquare$

Worked Example: Use the sequential criterion to show $f(x) = \sin(1/x)$ has no limit as $x \to 0$

Solution. Consider the sequences $x_n = 1/(2n\pi)$ and $y_n = 1/(2n\pi + \pi/2)$ . Both converge to $0$ . But $f(x_n) = \sin(2n\pi) = 0$ and $f(y_n) = \sin(2n\pi + \pi/2) = 1$ for all $n$ .

So $f(x_n) \to 0$ and $f(y_n) \to 1$ . By the sequential criterion, if $\lim_{x \to 0} f(x)$ existed, Both subsequences would converge to the same limit. Since they don”t, the limit does not exist. $\blacksquare$

Worked Example: Prove $f(x) = x \sin(1/x)$ (with $f(0) = 0$) is continuous everywhere

Solution. For $x \neq 0$ , $f$ is a product of continuous functions, hence continuous.

At $x = 0$ : let $\varepsilon > 0$ . Choose $\delta = \varepsilon$ . For $|x - 0| = |x| \lt \delta$ :

$|f(x) - f(0)| = |x \sin(1/x)| \leq |x| \lt \delta = \varepsilon$

So $f$ is continuous at $0$ . Since $f$ extends continuously from $(0, 1]$ to $[0, 1]$ The Heine-Cantor Theorem implies $f$ is uniformly continuous on $[0, 1]$ . $\blacksquare$

Worked Example: $\varepsilon$-$\delta$ .../1-number-and-algebra/3_proof-and-logic that $f(x) = \sin x$ is continuous at every $a \in \mathbb{R}$

Solution. We use the identity $|\sin u - \sin v| \leq |u - v|$ for all $u, v \in \mathbb{R}$ . (Proof: $|\sin u - \sin v| = 2|\cos((u+v)/2)\sin((u-v)/2)| \leq 2|\sin((u-v)/2)| \leq |u - v|$ Using $|\sin t| \leq |t|$ and $|\cos| \leq 1$ .)

Let $\varepsilon > 0$ and $a \in \mathbb{R}$ . Choose $\delta = \varepsilon$ . For $|x - a| \lt \delta$ :

$|\sin x - \sin a| \leq |x - a| \lt \delta = \varepsilon$

Since $\delta = \varepsilon$ works independently of $a$ , $\sin x$ is actually uniformly continuous On $\mathbb{R}$ . The same argument works for $\cos x$ . $\blacksquare$

Worked Example: $\varepsilon$-$\delta$ .../1-number-and-algebra/3_proof-and-logic that $f(x) = e^x$ is continuous at every $a \in \mathbb{R}$

Solution. We use the inequality $|e^u - e^v| \leq e^{\max(u,v)} |u - v|$ Which follows from the Mean Value Theorem applied to $e^t$ : $e^u - e^v = e^\xi (u - v)$ for some $\xi$ between $u$ and $v$ So $|e^u - e^v| = e^\xi |u - v| \leq e^{\max(u,v)} |u - v|$ .

Let $\varepsilon > 0$ and $a \in \mathbb{R}$ . Restrict to $|x - a| \lt 1$ So $x \lt a + 1$ and $e^{\max(x,a)} \leq e^{a+1}$ . Choose $\delta = \min(1, \varepsilon / e^{a+1})$ . For $|x - a| \lt \delta$ :

$|e^x - e^a| \leq e^{a+1} |x - a| \lt e^{a+1} \cdot \frac{\varepsilon}{e^{a+1}} = \varepsilon$

$\blacksquare$

If you get this wrong, revise: Section 4.2 (Continuity), Section 5.3 (Mean Value Theorem).

:::caution Common Pitfall Continuity on $(a, b)$ does not imply uniform continuity. The function $f(x) = 1/x$ on $(0, 1)$ is Continuous but not uniformly continuous. The Heine-Cantor theorem requires a closed and bounded Interval. Also, a function can be uniformly continuous on an unbounded domain (e.g., $f(x) = \sqrt{x}$ On $[0, \infty)$ ) --- boundedness of the domain is sufficient but not necessary.

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