Functions you can draw without lifting the pen. Hunting roots by bisection, a wobbly table you can always fix, and behaviour at infinity
A continuous function has no jumps, holes or blow-ups: its value at each point is exactly what the nearby values are heading for. That modest promise has surprising consequences, from finding roots to steadying a café table.
Drag a and b so that f has opposite signs at the ends, then step. Each step halves the bracket that must contain a root. Try 1/x to see what goes wrong without continuity.
The last column is log₁₀ of how many times the bracket has shrunk: digits gained, rising by 0.301 a row. The circles mark roots found by scanning when the endpoint signs agree. After about 40 halvings a bracket of width 2 is down to 2 × 10⁻¹², near the limit of double-precision arithmetic.
f is continuous at a when lim_(x → a) f(x) = f(a): the limit exists, f(a) exists, and they agree. The intermediate value theorem says that if f is continuous on [a, b] and N lies between f(a) and f(b), then f(c) = N for some c in [a, b]. With N = 0: a continuous function that changes sign has a root.
Bisection turns the theorem into a proof and an algorithm at once. If f(m) at the midpoint is not zero, it has the opposite sign to f at one end, so one half still brackets a sign change. The nested brackets shrink to a single point c; continuity forces f(c) to be the limit of values of both signs, so f(c) = 0. That last step needs the completeness of the real numbers: over the rationals, x² − 2 changes sign on [1, 2] with no root at all.
After n steps the bracket has width (b − a)/2ⁿ. Since log₁₀ 2 ≈ 0.301, every step gains about 0.3 decimal digits, slow but certain. Bolzano proved the theorem in 1817 in an attempt to put analysis on purely arithmetic foundations, without appeals to geometric intuition.
The theorem promises nothing when the signs agree: x² − 2 on [−2, 2] has two roots, x² + 2 none. And it needs continuity: 1/x is negative at −1 and positive at 1 but never zero, because it crosses from −∞ to +∞ instead of through 0.
A square table on uneven ground rocks on three feet. Turn it and watch the gap under the fourth foot change sign, or let bisection find an angle where all four touch.
The model: feet at the corners of a square, turned by θ about the vertical axis through the centre, at fixed horizontal positions (a tilted real table shifts its feet sideways very slightly, which is ignored here). Feet 1, 2 and 3 rest on the ground and fix a plane; foot 0 is on that plane, and g(θ) is its height above the ground. For a square, that plane meets foot 0’s corner at z₁ + z₃ − z₂, so g = z₁ + z₃ − z₂ − z₀, and a quarter turn relabels the feet and flips the sign. Yellow means foot 0 is in the air, red means it would be underground.
Let g(θ) be the gap under foot 0 when the table, turned by θ, rests on the other three. Turning by 90° moves every foot to its neighbour’s position, and for a square table that turns the gap into its negative: g(θ + 90°) = −g(θ). If g(0) is positive then g(90°) is negative, and since g changes continuously, the intermediate value theorem gives an angle in between with g = 0, where all four feet touch.
The argument is old mathematical folklore. Making it rigorous is subtler than the picture suggests. A real table tilts, its feet move sideways as it does, the ground might be too steep, and resting on three feet need not pick out a unique position. Bill Baritompa, Rainer Löwen, Burkard Polster and Marty Ross proved a careful version in “Mathematical table-turning revisited” (2007): if the ground never slopes by more than about 35°, a square table can be turned through less than 90° to rest on all four feet. The demo uses the simplified model described under it.
One caveat from the real world: the theorem promises a table that stands firm, not one that is level. On a slope all four feet can touch while the top tilts.
Widen the window up to a million units. Asymptotes appear, near-field detail vanishes, and in the race below every exponential eventually beats every power.
For large |x| only the leading terms matter: f(x) ≈ 2x² / x² = 2. So f(x) → 2 as x → ±∞: the horizontal asymptote y = 2.
At x = 5, f(x) = 2.42857, 0.429 from the asymptote. Near the origin the poles at x = ±2 dominate the picture; from far away only the horizontal line y = 2 is left.
numerator 2x² + 1; denominator x² − 4
The lower panel races six functions from x = 1 to x = 10 on log–log axes, where every power xᵖ is a straight line of slope p and eˣ curves away upward; the values listed are at the right edge. eˣ leads. ln x is still ahead of x⁰·¹, and stays ahead until x ≈ 3.4 × 10¹⁵: every positive power beats the logarithm eventually, however small. A typed function is checked for being a ratio of polynomials straight from its formula; other formulas are just plotted.
lim_(x → ∞) f(x) = L means f(x) can be made as close to L as we like by taking x large enough: for every ε > 0 there is an M with |f(x) − L| < ε whenever x > M. The line y = L is then a horizontal asymptote.
For a ratio of polynomials, divide top and bottom by the highest power of x in the denominator. Every term with a lower power tends to 0, leaving the ratio of the leading terms. Equal degrees give the horizontal asymptote y = (leading coefficient ratio); a lower top degree gives y = 0; a top degree one higher gives a slant asymptote, found by long division; higher still and f grows without bound.
Beyond polynomials there is a strict ranking of growth as x → ∞: ln x ≪ xᵖ ≪ eˣ for every p > 0. So xⁿ/eˣ → 0 for every n, and ln x/xᵖ → 0 for every p > 0, but the ranking can take a long time to show. eˣ/x¹⁰ falls to about 2 × 10⁻⁶ at x = 10 before the exponential takes over near x = 35.8, and x⁰·¹ only overtakes ln x near 3.4 × 10¹⁵. The two crossover points are related: both come from solving t = e^(t/10).
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