Grade 11

Continuous Functions

A function is continuous at a point when its graph passes through that point with no hole, jump, or break: formally, $\lim_{x \to x_0} f(x) = f(x_0)$ . Continuity turns out to guarantee two of the most useful facts in analysis — the Intermediate Value Theorem (a continuous function hits every value between $f(a)$ and $f(b)$ , the basis of the bisection root-finding algorithm) and the Extreme Value Theorem (a continuous function on a closed interval always attains a maximum and a minimum).

IntuitionDrawing a curve without lifting the pen

Take a pencil and draw the graph of $y=f(x)$ from left to right. If you never have to lift the pencil off the paper — no holes, no sudden jumps, no vertical asymptotes shooting off to infinity — the function is continuous on that stretch. The moment you must lift the pencil, the function has a discontinuity at that point. This picture is informal, but it points straight at the precise definition below: a function is continuous at $x_0$ exactly when its value $f(x_0)$ and its limiting behavior as $x\to x_0$ agree.

f(x) = x^3 - 3x + 1

. This curve never lifts the pen: it is continuous everywhere. Because

f(0) = 1 > 0

and

f(1) = -1 < 0

, the curve must cross the

x

-axis somewhere between

x=0

and

x=1

— this is exactly the Intermediate Value Theorem in action, and it is the same cubic used throughout this page.

SchoolContinuity at a point

Definition: Continuity at a point

Let $f$ be defined on an open interval containing $x_0$ . We say $f$ is **continuous at $x_0$ ** when three conditions all hold: (1) $f(x_0)$ is defined; (2) the limit $\lim_{x \to x_0} f(x)$ exists; (3) that limit equals $f(x_0)$ . Combined, these three conditions are written as a single equation, $\lim_{x \to x_0} f(x) = f(x_0)$ . A function is continuous on an open interval $(a;b)$ if it is continuous at every point of that interval, and continuous on a closed interval $[a;b]$ if in addition the one-sided limits at the endpoints match the function's values there.

\lim_{x \to x_0} f(x) = f(x_0)

Every polynomial, and every function built from polynomials, $\sin$ , $\cos$ , exponentials and roots by addition, multiplication, division (away from zeros of the denominator) and composition, is continuous at every point of its domain — this is why continuity is "automatic" for almost every formula you write down in school mathematics. Discontinuities only appear at isolated exceptional points: division by zero, a piecewise definition that does not match up, or a genuinely pathological construction.

\lim_{x \to x_0^-} f(x) \neq \lim_{x \to x_0^+} f(x)

Three ways continuity can fail at $x_0$
Type	Condition	Example at $x_0$
Removable	$\lim_{x \to x_0} f(x)$ exists but $\lim_{x \to x_0} f(x) \neq f(x_0)$ , or $f(x_0)$ is undefined.	$f(x) = \dfrac{x^2-1}{x-1}$ at $x_0=1$ : the limit is $2$ but $f(1)$ is undefined.
Jump	Both one-sided limits are finite but $\lim_{x \to x_0^-} f(x) \neq \lim_{x \to x_0^+} f(x)$ .	$f(x) = \lfloor x \rfloor$ at $x_0=1$ : left limit $0$ , right limit $1$ .
Infinite	At least one one-sided limit is $\pm\infty$ , so $\lim_{x \to x_0} f(x) = \pm\infty$ (no finite limit exists).	$f(x) = \dfrac{1}{x}$ at $x_0=0$ : both one-sided limits are infinite.

UndergraduateTwo pillars of continuous functions on an interval

Theorem: Intermediate Value Theorem (Bolzano–Cauchy)

If $f$ is continuous on $[a,b]$ and $k$ lies between $f(a)$ and $f(b)$ , then $\exists\, c \in (a,b)$ with $f(c)=k$ . In particular, if $f(a)$ and $f(b)$ have opposite signs, then $f(a)\cdot f(b) < 0 \ \Rightarrow\ \exists\, c \in (a,b): f(c) = 0$ .

Why is it true?

Think of $f$ as recording your altitude while hiking continuously from town $a$ (altitude $f(a)$ ) to town $b$ (altitude $f(b)$ ) along a trail with no teleporting. You cannot skip over any altitude strictly between the two towns' altitudes — at some instant you must have been at every intermediate height. A discontinuous function can skip values (think of a jump discontinuity), which is exactly why continuity is the essential hypothesis.

Proof

Assume without loss of generality the general case $f(a) < k < f(b)$ (the case $f(a) > k > f(b)$ is symmetric), and set $g(x) = f(x) - k$ , so $g$ is continuous, $g(a) < 0 < g(b)$ , and we must find $c$ with $g(c)=0$ . Set $a_0=a$ , $b_0=b$ .

Repeat the following bisection step for $n=0,1,2,\dots$ : compute the midpoint $m = \dfrac{a+b}{2}$ of $[a_n,b_n]$ . If $g(m)=0$ , stop — $c=m$ is found. Otherwise $g(m)$ has a definite sign; keep whichever half of $[a_n,b_n]$ has endpoints of opposite sign, i.e. set $[a_{n+1},b_{n+1}] = [a_n,m]$ if $g(a_n)g(m)<0$ , or $[a_{n+1},b_{n+1}]=[m,b_n]$ if $g(m)g(b_n)<0$ . In every case $g(a_{n+1})$ and $g(b_{n+1})$ keep opposite signs, i.e. $f(a_n)\cdot f(b_n) \le 0$ (with $a_n,b_n$ now roles of $g$ ), and the interval length halves: $b_n - a_n = \dfrac{b-a}{2^n}$ .

The sequence $(a_n)$ is increasing and bounded above by $b$ , and $(b_n)$ is decreasing and bounded below by $a$ , so both converge; since $b_n-a_n\to 0$ , they converge to the same limit, $a_n \to c,\quad b_n \to c$ , with $c\in[a,b]$ .

By continuity of $g$ , $g(a_n)\to g(c)$ and $g(b_n)\to g(c)$ . Because $g(a_n)$ and $g(b_n)$ have opposite signs at every stage (or one is already $0$ ), $g(a_n)g(b_n)\le 0$ for all $n$ ; passing to the limit gives $g(c)^2\le 0$ , i.e. $f(c)^2 \le 0 \ \Rightarrow\ f(c) = 0$ (applied to $g$ ). Hence $f(c) = g(c)+k = k$ , and $c\in(a,b)$ since $a_n<c<b_n$ strictly as soon as bisection has run at least one step without landing on an endpoint (and if it lands on $m$ at some step, that $m$ is the sought $c$ ).

Theorem: Extreme Value Theorem (Weierstrass)

If $f$ is continuous on the closed, bounded interval $[a,b]$ , then $\exists\, c, d \in [a,b]:\ f(c) \le f(x) \le f(d)\ \ \forall x \in [a,b]$ : that is, $f$ attains a maximum value $f(d)$ and a minimum value $f(c)$ somewhere on $[a,b]$ .

Why is it true?

A continuous function on a closed interval cannot "run away" to infinity (the interval is bounded and closed, so there is nowhere to escape to), and it cannot approach a highest value without ever quite reaching it either — an open interval like $(0,1)$ lets $f(x)=x$ get arbitrarily close to $1$ without ever attaining it, but including the endpoint $x=1$ in $[0,1]$ closes that loophole. Both the boundedness of the interval and its closedness (including endpoints) are essential.

Proof

**Step 1 — $f$ is bounded above on $[a,b]$ .** Suppose not: for every $n\in\mathbb{N}$ there is $x_n\in[a,b]$ with $f(x_n) > n$ . The sequence $(x_n)$ lies in the bounded interval $[a,b]$ , so by the Bolzano–Weierstrass theorem it has a subsequence $x_{n_k}\to c\in[a,b]$ . Continuity of $f$ gives $f(x_{n_k})\to f(c)$ , a single finite number, contradicting $f(x_{n_k}) > n_k \to \infty$ . Hence $f$ is bounded above; the symmetric argument (applied to $-f$ ) shows $f$ is bounded below.

Step 2 — the supremum is attained. By Step 1, $M=\sup_{[a,b]} f$ is a finite number. By definition of supremum, for every $n$ there is $y_n\in[a,b]$ with $M-\tfrac1n < f(y_n) \le M$ . Again by Bolzano–Weierstrass, $(y_n)$ has a subsequence $y_{n_k}\to d\in[a,b]$ . Squeezing $M-\tfrac{1}{n_k} < f(y_{n_k}) \le M$ as $k\to\infty$ gives $f(y_{n_k})\to M$ ; but continuity also gives $f(y_{n_k})\to f(d)$ , so $f(d)=M$ . Thus the maximum value $M$ is attained at $x=d$ .

Step 3 — the minimum is attained. Apply Steps 1–2 to $-f$ (also continuous on $[a,b]$ ): $-f$ attains a maximum at some $c\in[a,b]$ , which means $f$ attains its minimum at that same $c$ . This gives both $c$ and $d$ as required.

UndergraduateReal-World Applications and Worked Examples

The Intermediate Value Theorem is not just an existence statement — it is the mathematical backbone of the bisection method, one of the oldest and most reliable numerical root-finding algorithms in engineering and computer science: given $f(a)$ and $f(b)$ of opposite sign, repeatedly halving the interval (exactly as in the proof above) traps a root to any desired precision, and every step is guaranteed to succeed because $f$ never "jumps over" zero. The same theorem also justifies equilibrium arguments in physics: whenever a continuous quantity (temperature, pressure, concentration) is compared between two states or two locations, IVT guarantees an intermediate state or location where the two sides balance exactly.

Example: Locating a root with the bisection method

Show that $f(x) = x^3 - 3x + 1$ has a root in $(0,1)$ , then use the bisection method to locate it to two decimal places.

Solution

Since $f$ is a polynomial it is continuous everywhere, and $f(0) = 1 > 0$ while $f(1) = -1 < 0$ , so $f(0)f(1)<0$ ; by IVT there is a root $c\in(0,1)$ .

Bisect: $m=0.5$ , $f(0.5)=0.125-1.5+1=-0.375<0$ . Since $f(0)>0>f(0.5)$ , the root lies in $(0, 0.5)$ .

Bisect again: $m=0.25$ , $f(0.25)=0.015625-0.75+1=0.265625>0$ . Since $f(0.25)>0>f(0.5)$ , the root lies in $(0.25, 0.5)$ .

Continuing — $m=0.375$ gives $f(0.375)\approx -0.072<0$ (root in $(0.25,0.375)$ ), then $m=0.3125$ gives $f(0.3125)\approx 0.093>0$ (root in $(0.3125,0.375)$ ), then $m=0.34375$ gives $f(0.34375)\approx 0.009>0$ (root in $(0.34375,0.375)$ ) — the interval has shrunk to length under $0.031$ and both endpoints round to $0.34$ – $0.38$ ; one more step confirms the root rounds to $c \approx 0{,}35$ , matching the widget's marked crossing near $x\approx 0.35$ .

Example: Two antipodal points at the same temperature

Around a circular running track of circumference $2\pi$ , the ground temperature $T(\theta)$ (as a function of the angle $\theta$ ) varies continuously. Show that at some instant there are two diametrically opposite points on the track with exactly the same temperature.

Solution

Define $g(x) = T(x) - T(x+\pi)$ for $x\in[0,\pi]$ , comparing the temperature at angle $x$ with the temperature at the diametrically opposite angle $x+\pi$ . Since $T$ is continuous, $g$ is continuous on $[0,\pi]$ .

Evaluate at the endpoints: $g(0) = T(0)-T(\pi)$ , and $g(\pi) = T(\pi) - T(2\pi) = T(\pi) - T(0) = -g(0)$ — that is, $g(\pi) = -g(0)$ .

If $g(0)=0$ , then $T(0)=T(\pi)$ already, and $x=0,\pi$ are the two antipodal points sought. Otherwise $g(0)$ and $g(\pi)=-g(0)$ have strictly opposite signs, so $g(0)g(\pi)<0$ ; by the Intermediate Value Theorem there is $c\in(0,\pi)$ with $g(c)=0$ , i.e. $T(c) = T(c+\pi)$ . In either case, the points at angles $c$ and $c+\pi$ have identical temperature — this is exactly the "thermal equilibrium" flavor of the theorem: continuity forces two opposing readings to cross.

ResearchContinuity at the research frontier

Research frontier as of 2026

As of 2026, two directions extend continuity and its bisection-style existence arguments well beyond hand computation. (1) Certified numerical computing. Interval arithmetic tracks a guaranteed enclosing interval for every intermediate computation instead of a single floating-point number, so a computer can output a mathematically rigorous, machine-checked proof that a root exists in a tiny interval — this technique underlies computer-assisted proofs such as the resolution of the double-bubble conjecture and parts of the proof of the Kepler conjecture. (2) Higher-dimensional generalizations. The circular "two antipodal points" argument above is a one-dimensional shadow of the Borsuk–Ulam theorem (for every continuous map from a sphere to $\mathbb{R}^n$ , some pair of antipodal points has equal image), a cornerstone of algebraic topology with modern applications ranging from fair-division algorithms (splitting a sandwich or an election district evenly) to combinatorics and theoretical computer science.

Let $f(x) = \dfrac{x^2-9}{x-3}$ for $x\neq 3$ and $f(3)=5$ . Is $f$ continuous at $x_0=3$ ?

$f(x) = \lfloor x \rfloor$ has, at every integer $x_0$ , a discontinuity of which type?

Does $f(x) = x^3 - 3x + 1$ have a root in the interval $(1,2)$ ?

Applying the bisection method to $f(x) = x^3 - 3x + 1$ starting from $[0,1]$ , what is the new bracketing interval after the first step?

References

James Stewart (2015). Calculus: Early Transcendentals
Walter Rudin (1976). Principles of Mathematical Analysis
Francis Edward Su (1997). A Borsuk-Ulam Equivalent that Directly Implies the Sperner Lemma