What is continuity in simple words?

A function is continuous if you can draw its graph without lifting your pencil. Small changes in x give small changes in f(x).

What does the intermediate value theorem say?

If f is continuous on [a, b], then f takes every value between f(a) and f(b) at least once inside [a, b].

How do I show an equation has a solution?

Write it as f(x) = 0, check f is continuous, find two points where f has opposite signs, then quote the intermediate value theorem.

Continuity and Intermediate Value Theorem Explained

A function is continuous at a point if its limit there equals its value. A continuous function on an interval has no jumps: it cannot skip a value between two of its values (intermediate value theorem). If it is also strictly monotone, it takes each such value exactly once.

Continuity at a point and on an interval

A function f is continuous at a when three things are true: f(a) exists, the limit of f(x) as x goes to a exists, and the two are equal: lim f(x) = f(a).

In simple words: as x creeps up on a from the left and from the right, f(x) creeps up on the same number f(a). There is no hole, no jump, no blow-up.

f is continuous on an interval if it is continuous at every point of it. At the end points of a closed interval [a, b] we only use the one side that exists.

Good news: polynomials, sin, cos, exp are continuous everywhere. A fraction like 1/x is continuous everywhere it is defined, but it is not defined at 0, so there it breaks. Sums, products, and compositions of continuous functions are continuous.

Differentiable implies continuous

If f has a derivative at a, then f is continuous at a. A tangent line can only exist on a curve that has no break. So differentiable ⇒ continuous.

The other way is false. The function |x| is continuous at 0 (no break) but has a sharp corner, so it has no derivative at 0. Continuous is a weaker idea than differentiable.

The image of a convergent sequence

Take a sequence uₙ that converges to L. If f is continuous at L, then the sequence f(uₙ) converges to f(L). We write: lim f(uₙ) = f(lim uₙ).

Example: uₙ = 2 + 1/n goes to 2. For f(x) = x², f(uₙ) goes to f(2) = 4. This lets us move the limit inside a continuous function. If uₙ₊₁ = f(uₙ) and uₙ converges to L, then L = f(L): the limit is a fixed point.

Intermediate value theorem (IVT)

Let f be continuous on [a, b]. If k is any number between f(a) and f(b), then there is at least one c in [a, b] with f(c) = k.

Use it to show a root exists: if f(a) and f(b) have opposite signs, take k = 0, so f has a zero between a and b. Example: f(x) = x² − 5 has f(2) = −1 and f(3) = 4, so x² = 5 has a solution between 2 and 3.

Both conditions matter: f must be continuous on the whole closed interval. On 1/x over [−1, 1] the signs change but there is no root, because 1/x breaks at 0.

Bisection: keep halving the interval, keeping the half where the sign changes, to trap the root as tightly as you like.

The strictly monotone case

If f is continuous and strictly increasing (or strictly decreasing) on [a, b], then each value k between f(a) and f(b) is reached exactly once. Monotone gives "at most one", IVT gives "at least one".

This is how we prove that an equation f(x) = k has a unique solution, for example x³ + x = 5 (the derivative 3x² + 1 is positive, so f is strictly increasing). It is also why inverse functions of continuous strictly monotone functions exist.

Key formulas and definitions

Continuous at a: lim(x→a) f(x) = f(a)
Differentiable at a ⇒ continuous at a
If uₙ → L and f continuous at L: f(uₙ) → f(L)
IVT: f continuous on [a, b], k between f(a) and f(b) ⇒ f(c) = k for some c in [a, b]
Root test: f(a) × f(b) < 0 ⇒ a root in (a, b)
Bisection midpoint: m = (a + b) / 2

Worked examples

1. Is f(x) = 3x² − 2x continuous at x = 4? Find f(4).

f is a polynomial, so it is continuous everywhere. lim f(x) = f(4) = 3(16) − 8 = 40.

2. f(x) = x + 1 for x < 2 and f(x) = ax − 1 for x ≥ 2. Find a so that f is continuous at 2.

Left limit = 2 + 1 = 3. Value at 2 = 2a − 1. Set equal: 2a − 1 = 3, so a = 2.

3. f(x) = (x² − 4)/(x − 2) for x ≠ 2. What value of f(2) makes f continuous at 2?

For x ≠ 2, f(x) = (x − 2)(x + 2)/(x − 2) = x + 2. The limit at 2 is 4, so define f(2) = 4.

4. Show that x³ − x − 1 = 0 has a root between 1 and 2.

f(x) = x³ − x − 1 is a polynomial, so continuous. f(1) = −1 < 0 and f(2) = 5 > 0. By the IVT there is c in (1, 2) with f(c) = 0.

5. Use one bisection step for x² − 2 = 0 on [1, 2]. Which half has the root?

Midpoint 1.5: f(1.5) = 2.25 − 2 = 0.25 > 0. f(1) = −1 < 0. The sign changes between 1 and 1.5, so the root is in [1, 1.5].

6. Show x³ + x = 5 has exactly one solution.

f(x) = x³ + x − 5 is continuous. f(1) = −3 and f(2) = 5, so there is a root in (1, 2) by the IVT. f′(x) = 3x² + 1 > 0, so f is strictly increasing and can cross zero only once. Exactly one solution.

Common mistakes

Thinking "limit exists" is enough. The limit must also equal f(a), and f(a) must exist.
Using the IVT on an interval where f is not continuous (like 1/x over [−1, 1]).
Saying continuous means differentiable. |x| is continuous at 0 but has no derivative there.
Believing the IVT gives the exact root or says there is only one. It only says at least one exists.

Practice quiz

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Continuity and the Intermediate Value Theorem

🎬 Step-by-step story

🤔 Common doubts, cleared

Does "continuous" really mean I can draw it without lifting the pencil?

What goes wrong at the broken point?

Why must the green line touch a smooth curve?

Why does the line miss the broken curve?

If the curve crosses the line, is it crossed only once?

Is the IVT true if f is continuous but I use an open interval?