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1

The definition, as a game

Two sliders, one promise

Our running example is the robot from Part 1, whose position is $s(t) = \tfrac{1}{3}t^3 - \tfrac{3}{2}t^2 + 2t$ metres at time $t$ seconds. Pick a time $a$ and set $L = s(a)$. The claim that the limit of $s$ at $a$ is $L$, written $\lim_{t\to a} s(t) = L$, is not "it gets closer and closer". It is this precise statement:

$$ \forall \varepsilon > 0 \;\; \exists \delta > 0 \quad\text{such that}\quad 0 < |t - a| < \delta \;\Longrightarrow\; |s(t) - L| < \varepsilon. $$

Read it as a game. You are the adversary: you name a tolerance $\varepsilon$ — how close to $L$ the output must be. The function answers with a window radius $\delta$ — how close to $a$ the input must stay. The order of the quantifiers is the content: the answer $\delta$ is chosen after seeing $\varepsilon$, and is allowed to depend on it. The strict inequality $0 < |t-a|$ excludes $t = a$ itself, so the limit is about the approach and never about the value $s(a)$.

Drag the three sliders. The blue band is $L \pm \varepsilon$; the pink window is $a \pm \delta$. The curve is drawn green inside the window wherever it stays in the band, red wherever it escapes.

The ε-band around $L = s(a)$, the δ-window around $a$, and the curve coloured by whether it stays inside the band on that window.

2

One side at a time

Left and right approaches

A two-sided limit ignores which direction the input comes from. Split the approach in two and the notation grows a sign:

$$ \lim_{t\to a^-} s(t) \quad\text{and}\quad \lim_{t\to a^+} s(t). $$

The two-sided limit exists exactly when both one-sided limits exist and agree. Slide the left and right probes toward the point $a = 1$; the readout reports the height of the curve at each probe and the values the two sides are heading for. Switch to the step function and the two sides settle on different heights — that is the definition of a limit that does not exist.

Probes at $a-h$ and $a+h$ with guide lines to the value axis. The dashed vertical is the point $a$ itself.

3

Where limits fail

Four ways to break the promise

The definition can fail for four different reasons, and the picture tells them apart. A removable hole has a perfectly good limit that the function simply does not take. A jump has two one-sided limits that disagree. A vertical asymptote is unbounded, so no finite $L$ can satisfy the band. And an oscillation never settles: however thin the window, the curve keeps sweeping across the band.

Pick a function and move the band and window. The verdict names which clause of the definition is violated.

The ε-band sits around a candidate $L$; the δ-window around $x = 1$. Green means the curve stays in the band there, red means it escapes.

4

Continuity in three checks

Defined, limit exists, equal

A function is continuous at $a$ when the value and the limit are the same number. Unpacking that gives three conditions, checked in order:

$$ \underbrace{f(a)\ \text{is defined}}_{1}\qquad \underbrace{\lim_{x\to a} f(x)\ \text{exists}}_{2}\qquad \underbrace{\lim_{x\to a} f(x) = f(a)}_{3}. $$

Condition 2 is the work: both one-sided limits must exist and agree. Drag the probe along the curve, or pick a function and slide it onto the bad point. The checklist turns green only when all three hold; the first failure names what went wrong.

The probe $a$ can be dragged along the curve or set with the slider. The vertical line marks the point under test.

5

Why "closer and closer" is not enough

The quantifiers are doing real work

The informal phrase hides two mistakes. First, it lets the input choose the path: a function can be made to look convergent if you only sample along a sequence you picked. Second, it lets $\delta$ depend on the wrong thing. Consider $f(x) = \sin(1/x)$ near $0$. Along the sequence $x_n = 1/(\pi/2 + 2\pi n)$ the value is exactly $1$ for every $n$, and along $x_n' = 1/(3\pi/2 + 2\pi n)$ it is exactly $-1$. Both sequences march toward $0$, yet the values do not settle on any single number — between them the curve sweeps the whole interval $[-1, 1]$.

⚠️ "Closer" along some path is not "close" along every path. The quantifier order repairs this: for every $\varepsilon$ there must be one $\delta$ that works for all $x$ in the window — not merely for the points you chose to inspect. Swapping the order to "$\exists\delta\;\forall\varepsilon$" is a different, usually false claim: it would mean a single window whose values are already within every tolerance, which forces $f$ to be constant there.
6

Where this shows up

The next page is built on it

The derivative of Part 3 is exactly the limit $\lim_{h\to 0}\frac{s(a+h) - s(a)}{h}$, and the game you just played is what makes that limit meaningful: for each tolerance on the slope, there must be a window of step sizes that delivers it. Every later appearance of calculus — the gradient of a loss, the error term in a Taylor polynomial, the convergence test for a solver — is this same quantifier pattern wearing different clothes. The smooth-versus-corner distinction from Part 3 is also the continuity question of Step 4: differentiable implies continuous, but not the reverse.

7

Notation and laws to carry forward

Notation / lawReads asWhere it comes up
limx→a f(x) = LThe two-sided limit, defined by the ε–δ statementEvery definition in this volume
∀ε>0 ∃δ>0 : 0<|x−a|<δ ⇒ |f(x)−L|<εThe definition; δ may depend on ε, never on xStep 1
limx→a⁻ f(x), limx→a⁺ f(x)One-sided limits; the two-sided limit exists iff both exist and agreeStep 2, jumps
lim(f+g) = lim f + lim gLimits split over sumsLimit algebra, every proof
lim(f·g) = (lim f)(lim g)Limits split over productsLimit algebra
lim(f/g) = (lim f)/(lim g) if lim g ≠ 0Quotients split only when the denominator does not vanishStep 3, asymptotes
g ≤ f ≤ h and lim g = lim h = L ⇒ lim f = LThe squeeze theorem: pin an unknown function between two known onesStep 5, oscillation
f continuous at a ⟺ defined, limit exists, equalThe three checksStep 4; Part 3's differentiability
8

Further reading

9

Check your understanding

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