Discover the epsilon-delta definition and properties of continuous functions
A continuous function has no jumps: nudge the input a little and the output moves a little. These demos make that idea precise and show what it buys you.
Pick a function and a point a, then shrink epsilon. Watch how small delta must get to keep the graph inside the band.
f is continuous at a if: for every ε > 0, there exists δ > 0 such that
|x - a| < δ ⟹ |f(x) - f(a)| < ε
If the curve stays within the green ε-band whenever x is in the red δ-interval, then δ works for this ε.
Continuity captures the intuitive idea that a function has "no jumps" -- small changes in input produce small changes in output. The rigorous epsilon-delta definition states: for every epsilon > 0, there exists delta > 0 such that |x - a| < delta implies |f(x) - f(a)| < epsilon.
Explore the formal definition interactively. Adjust epsilon to see how delta must respond to keep the function values within the target band.
Key insight: The order of quantifiers matters. "For every epsilon there exists delta" means the choice of delta can depend on epsilon (and on the point), but epsilon is chosen first by an adversary.
Select a removable, jump, infinite, or essential discontinuity, and turn on the limit lines to compare the left and right limits.
At x = 1:
Left limit: 2
Right limit: 2
f(1): undefined
The limit exists but f(a) is undefined or ≠ limit. Can be "fixed" by redefining f(a).
This function: Removable discontinuity at x = 1 (equals x + 1 elsewhere)
Not all discontinuities are created equal. Explore removable, jump, infinite, and essential discontinuities with visual examples. Understanding the classification helps determine which functions remain integrable.
Key insight: Removable and jump discontinuities are "tame" -- the function has well-defined one-sided limits. Essential discontinuities like sin(1/x) near zero are genuinely wild, with no limit at all.
Choose a function and interval, then press Step or Play to run bisection. Each halving keeps the crossing point trapped.
✓ IVT applies: f(a) and f(b) are on opposite sides of the target!
If f is continuous on [a, b] and y is any value between f(a) and f(b), then there exists some c ∈ (a, b) such that f(c) = y.
The bisection method exploits this: repeatedly halve the interval, keeping the half where the sign change occurs.
If a continuous function takes values f(a) and f(b) at the endpoints of an interval, it must take every value between them. Watch the bisection method find roots using this principle.
Key insight: The IVT requires both continuity and completeness of the real numbers. It fails for continuous functions on the rationals, showing again that completeness is essential.
Move the sample point and watch delta change. Compare 1/x near 0, where delta collapses, with the functions on closed intervals, where one delta works everywhere.
The same δ = 0.3030 works for all points in the interval.
Pointwise continuity: For each point x and ε > 0, ∃ δ > 0 (δ may depend on both x and ε)
Uniform continuity: For each ε > 0, ∃ δ > 0 that works for ALL points (δ depends only on ε, not on x)
Key theorem: A continuous function on a closed bounded interval [a, b] is always uniformly continuous.
Pointwise continuity allows delta to depend on the point x. Uniform continuity requires the same delta to work everywhere. Compare these concepts visually and see why uniform continuity is automatic on closed bounded intervals.
Key insight: The Heine-Cantor theorem guarantees that every continuous function on a closed bounded interval is uniformly continuous. This upgrades pointwise control to global control, essential for proving integrability.
Pick a function and drag the interval endpoints. The maximum and minimum are always reached somewhere in [a, b].
If f is continuous on a closed bounded interval [a, b], then f attains both a maximum and minimum value on that interval.
That is, there exist points c, d ∈ [a, b] such that:
f(c) ≤ f(x) ≤ f(d) for all x ∈ [a, b]
Key insight: The function must be continuous on a closed and bounded interval. Open intervals or discontinuous functions may fail to attain extrema.
A continuous function on a closed bounded interval always attains its maximum and minimum values. Both closedness and boundedness of the interval are essential -- removing either hypothesis allows the conclusion to fail.
Key insight: The EVT follows from the Bolzano-Weierstrass theorem applied to a maximizing sequence. It guarantees that optimization problems on compact domains always have solutions.
Spot a mistake or have an idea?