📊 Section 1 — Interactive Simulation
Pick a function and a point a. The simulation classifies the discontinuity (if any) by checking three numbers: the left limit, the right limit, and $f(a)$. Drag to pan; pinch / wheel to zoom.
Animation
Function Preset
Parameters
Display Options
Tips
• The simulation evaluates the function on a punctured neighbourhood of a from both sides and compares to $f(a)$.
• Try sliding a through a discontinuity — watch the classification badge flip.
• In ε-δ Continuity mode the test is $\abs{f(x)-f(a)}<\eps$ (no puncture — the value at $a$ matters here).
🌱 Section 2 — The Idea, Step by Step
From an everyday "unbroken road" picture up to the full $\forall\eps\,\exists\delta$ statement — one gentle climb, then three things to try in the simulation above.
Start — the unbroken-road picture
Picture driving along a road whose shape traces the graph of a function. Where the road runs smoothly under your wheels — no gap to fall into, no sudden step up or down, no ramp that shoots off to the sky — the function is continuous there. A discontinuity is any spot you'd have to jump, drop, or teleport across to keep going. There are exactly four ways a road can break: a single missing paving stone you could just drop back in (removable), a kerb where the surface steps to a new height (jump), a ramp that rockets upward without end (infinite), and a stretch that shudders up and down too fast to ever settle (essential).
Build — the three numbers that decide it
To test continuity at a point $a$ you only ever compare three numbers: the value the graph approaches from the left, $f(a^-)$; the value it approaches from the right, $f(a^+)$; and the value actually plotted at the point, $f(a)$. The function is continuous at $a$ exactly when all three agree. Miss on any one and you have a discontinuity.
One concrete case: take $f(x)=\dfrac{x^2-1}{x-1}$ at $a=1$. Cancel the common factor — $f(x)=\dfrac{(x-1)(x+1)}{x-1}=x+1$ for every $x\neq 1$ — so both sides approach $f(1^-)=f(1^+)=2$. But $f(1)$ itself is $0/0$, undefined. Two of the three numbers agree, the third is simply missing, so the graph has a single removable hole at the point $(1,2)$. Fill it by declaring $f(1):=2$ and the road is whole again.
Deepen — the precise statement and the sliders
The rigorous version drops the talk of roads: $f$ is continuous at $a$ iff $\forall\,\eps>0\ \exists\,\delta>0$ such that $\abs{x-a}<\delta \Rightarrow \abs{f(x)-f(a)}<\eps$. Unlike a limit there is no puncture — the point $x=a$ is included, because here the value at $a$ is precisely what the nearby values must match. In the simulation the $a$ slider chooses the test point, the $\eps$ slider sets the height of the band you demand the output stay within, and the $\delta$ slider sets the width of the window around $a$. The readouts report $f(a^-)$, $f(a^+)$, $f(a)$ and the jump size, while the classification badge flips between continuous, removable, jump, infinite, and essential as you slide.
Try this in the sim above
① Choose f(x) = x² and drag the point $a$ across the screen. The badge stays green Continuous everywhere, because the left value, the right value, and $f(a)$ always coincide — a smooth, unbroken road.
② Choose f(x) = sign(x) at $a=0$. The left limit is $-1$, the right limit is $+1$, so the jump readout reads $2$ and the badge turns red Jump. Shrink $\delta$ as small as you like — the window still straddles the step, so it can never be patched.
③ Choose f(x) = sin(1/(x−0.5)) at $a=0.5$ and zoom in with the wheel. The curve never settles, the one-sided limits "oscillate," and you get the pink Essential badge — the one kind of break you can never fill in.
📐 Section 3 — Definition & Classification Theorem
Three equivalent definitions of continuity, the symbol table, the classification of discontinuities, a proof, and a worked example.
$f : D \subseteq \R \to \R$ is continuous at $a \in D$ iff any (equivalently, all) of the following hold:
(C1) Limit form: $\displaystyle\lim_{x \to a} f(x) = f(a).$
(C2) ε-δ form: $\forall\,\eps>0\ \exists\,\delta>0:\ \abs{x-a}<\delta\ \Longrightarrow\ \abs{f(x)-f(a)}<\eps.$
(C3) Heine form (sequential): $\forall\{x_n\}\subset D,\ x_n\to a\ \Longrightarrow\ f(x_n)\to f(a).$
Note: in (C2) the puncture $0<\abs{x-a}$ is dropped — for continuity, $x=a$ is included (and trivially passes).
Symbol Table
| Symbol | Meaning | Type / Domain |
|---|---|---|
| $f$ | The function under study | $D \to \R$ |
| $D$ | Domain | Subset of $\R$, must contain $a$ |
| $a$ | Point of interest | Real, $a \in D$ |
| $f(a^-)$ | Left limit $\lim_{x\to a^-}f(x)$ | Real or $\pm\infty$ or DNE |
| $f(a^+)$ | Right limit $\lim_{x\to a^+}f(x)$ | Real or $\pm\infty$ or DNE |
| $\eps,\delta$ | Output / input tolerances | Both $>0$, $\delta=\delta(\eps)$ |
| $\omega(\delta)$ | Modulus of continuity $\sup\{\abs{f(x)-f(a)}:\abs{x-a}<\delta\}$ | Non-decreasing in $\delta$ |
Classification of Discontinuities
Suppose $f$ is not continuous at $a$. Examine $f(a^-),\, f(a^+),\, f(a)$:
Proof — (C1) ⇔ (C3)
Simulation ↔ Symbol Mapping
slider a | point $a$ being tested for continuity |
slider ε | output tolerance $\eps$ (vertical band around $f(a)$) |
slider δ | input radius $\delta$ around $a$ (horizontal strip — not punctured) |
readout f(a−) | $\displaystyle\lim_{h\to 0^+}f(a-h)$, computed by sampling $h\in\{0.5,0.25,\ldots,10^{-9}\}$ |
readout f(a+) | $\displaystyle\lim_{h\to 0^+}f(a+h)$, same scheme |
readout jump | $\abs{f(a^+)-f(a^-)}$; zero ⇒ left=right ⇒ candidate for continuity or removable |
readout continuous at a? | true iff both one-sided limits equal $f(a)$, all finite |
readout classification | continuous / removable / jump / infinite / essential — by the tree above |
graph ω(δ) | modulus of continuity — vanishes at 0 iff $f$ continuous at $a$ |
Worked Example — Classify and Patch
Let $\displaystyle f(x)=\frac{x^2-1}{x-1}$. Is $f$ continuous at $a=1$? If not, classify and (if possible) patch.
Domain: $f$ is undefined at $x=1$ (division by zero), so $a=1\notin D$. Already, $f$ cannot be continuous there in the strict sense — there's no $f(1)$.
Compute one-sided limits: for $x\neq 1$, $f(x)=\dfrac{(x-1)(x+1)}{x-1}=x+1$. So $f(1^-)=f(1^+)=2$.
Classification: both one-sided limits equal $L=2$, finite. Either way (whether $f(1)$ is undefined or unequal to $2$), this is a Removable discontinuity.
Patch: define $\tilde{f}(1):=2$. Then $\tilde{f}$ agrees with $f$ everywhere $f$ is defined and is continuous at $1$. $\boxed{\tilde{f}(x) = x+1.}$
❓ Section 4 — Frequently Asked Questions
Real questions Bangladeshi students ask about continuity. Tap to expand.
That intuition works for elementary functions on intervals but fails badly elsewhere. The Weierstrass function is continuous everywhere yet differentiable nowhere — drawing it requires infinite zoom-in detail; you can't physically draw it, but it is continuous. Conversely, a function defined on $\mathbb{Q}$ alone may be "continuous" in the formal sense even though its graph is a dust of disconnected points. The pen-lifting picture is a good warm-up, not a definition.
Key takeaway: continuity is a local ε-δ property — the pen test is a heuristic, not a definition.It samples $f$ at $a-h$ and $a+h$ for $h \in \{0.5, 0.25, \ldots, 10^{-9}\}$ to estimate the one-sided limits $f(a^-), f(a^+)$. It checks: (i) do these stabilize to finite values? (ii) do they agree? (iii) does that common value equal $f(a)$? If both stabilize and agree but differ from $f(a)$ (or $f(a)$ is undefined), it's removable. If they stabilize but disagree, it's jump. If one or both grow without bound, it's infinite. If they refuse to stabilize at all (the values keep oscillating as $h$ shrinks), it's essential.
Key takeaway: classification = (do limits exist?) + (are they equal?) + (do they match $f(a)$?).Step inputs in control systems (the unit-step / Heaviside function $H(t)$) are jump discontinuities — they drive the entire study of impulse response and convolution. Phase transitions in physics (water boiling, magnetization) appear as jump discontinuities of order parameters as a function of temperature. In programming, every if/else branch creates a piecewise function; if the branches don't agree at the boundary, you have a numerical discontinuity that can wreck optimization (gradient descent loses meaning at jumps). Shock waves in fluid dynamics are propagating jump discontinuities of pressure and density.
Yes — this is Thomae's function. Define $f(x)=1/q$ if $x=p/q$ in lowest terms with $q>0$, and $f(x)=0$ if $x$ is irrational, $f(0)=0$. Astonishingly, $f$ is continuous at every irrational and discontinuous at every rational. The reason: near any irrational $a$, all rationals with small denominator are far away, so $f$ values near $a$ are tiny — matching $f(a)=0$. By contrast, near any rational $p/q$, irrationals (where $f=0$) are dense, so $f$ keeps jumping back to $0$ while $f(p/q)=1/q\neq 0$. There is no function continuous on the rationals and discontinuous on the irrationals (a Baire-category result).
Key takeaway: the set of discontinuities of any function is always an $F_\sigma$ set — that's a deep restriction.It samples a geometric sequence $h_k = h_0 \cdot 2^{-k}$ for $k=0,1,\ldots,K$ (typically $K=30$, so $h$ shrinks from $0.5$ to $\approx 5\times 10^{-10}$). It computes $y_k = f(a\pm h_k)$ and checks whether $|y_K - y_{K-1}| < \tau$ for tolerance $\tau \approx 10^{-7}$. Cost: $O(K)$ function evaluations per side. Failure modes: (1) for $\sin(1/(x-a))$, $y_k$ oscillates without converging — flagged as "essential". (2) For $1/(x-a)^2$, $y_k$ grows like $4^k$ — flagged as "infinite" once $|y_k|$ exceeds $10^{12}$. (3) For Lipschitz $f$ near $a$, convergence is geometric so 30 steps is overkill — but cheap.
Key takeaway: numerical limits = "does $y_k$ stabilize as $h_k \to 0$ geometrically?" — $O(\log_2(1/\tau))$ work.This is the topological definition. Re-read (C2): $\abs{x-a}<\delta \Rightarrow \abs{f(x)-f(a)}<\eps$ says the preimage of $(f(a)-\eps,f(a)+\eps)$ contains $(a-\delta,a+\delta)$, an open interval around $a$. So for every open ball around $f(a)$, the preimage is open. Generalizing: $f$ is continuous on $D$ iff for every open set $U\subseteq\R$, $f^{-1}(U)$ is open in $D$. This works in arbitrary topological spaces — no metric needed. It also gives clean proofs: composition of continuous functions is continuous (preimage of open is open, applied twice), and continuous image of compact is compact (a topological one-liner).
Key takeaway: continuity is the morphism of topological spaces — ε-δ is just the metric expression of "preimage of open is open"."Continuous at $a$" is a local statement about one point. "Continuous on $[a,b]$" means continuous at every point of the interval (with one-sided continuity at the endpoints). The latter unlocks the heavy machinery: the Extreme Value Theorem (max/min attained), Intermediate Value Theorem (every value in between is hit), and uniform continuity (one $\delta$ works for the whole interval, not a different $\delta$ per point). On open intervals or on $\R$, continuity gives less — pointwise continuity does not imply uniform continuity (think of $f(x)=1/x$ on $(0,1)$).
Key takeaway: "continuous on a closed bounded interval" is what calculus's big theorems actually need.⚠️ Section 5 — Misconceptions & Common Errors
Three conceptual misconceptions about continuity, plus three procedural errors.