Calculus I › Limits › full formula sheet
Say it: “f of a is defined, the limit as x approaches a of f of x exists, and the limit equals f of a”
Continuity at a point
Three checks that answer: can I just plug in?
On this page “continuous at a” means all three conditions hold. Each condition has its own favorite counterexample — learn all three and you can diagnose any discontinuity on sight.
Notation in this lesson
- f(a)
- the function's value AT the point a
- limx→a f(x)
- the limit as x approaches a — nearby values, not the value at a
- (1)(2)(3)
- defined at a; limit exists at a; the two are equal
Where it comes from
Every “just plug in” you have ever done was a bet that the function is continuous at that point. Continuity is the property that makes plugging in legal — and it decomposes into three separately testable conditions. The fastest way to feel each one is to watch it fail:
The intuition: continuity means “the limit is what you’d get by plugging in” — no holes, no jumps, no misplaced points. Draw-without-lifting-the-pen is the cartoon; the three conditions are the testable version. And the payoff is huge: every limit law you learned preserves continuity, which is why polynomials — built from x and constants via sums, products, and powers — are continuous everywhere.
Derivation
Why is “plug in” legal for polynomials? Because continuity is preserved by the limit laws. Key steps:
Key steps shown; the argument above is complete. This is the deep reason the whole “evaluate limits” enterprise works: the functions you meet daily are continuous at the points you care about, so condition (3) — lim f(x) = f(a) — turns every limit into an evaluation. Discontinuities are the exceptions, which is why each deserves a name.
How to use it
The procedure — testing continuity at a, every time:
- Is f(a) defined? If no → discontinuous (removable hole, or worse). Fastest failure mode — check it first.
- Does lim(x→a) f(x) exist? Check both one-sided limits. If they disagree → jump discontinuity. If infinite or oscillating → infinite/essential discontinuity.
- Does the limit equal f(a)? If no → removable discontinuity (the point exists at the wrong height).
- All three yes? Continuous at a. Plug in freely.
The discontinuity zoo
Removable (hole or misplaced point): limit exists, but f(a) is missing or wrong — fixable by redefining one point. Jump: one-sided limits exist but disagree — not fixable by one point. Infinite: a one-sided limit is ±infinity (vertical asymptote). Oscillating: sin(1/x) at 0 — no one-sided limit at all.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — continuous: f(x) = x² − 3x + 1 at a = 2
- (1) Defined? f(2) = 4 − 6 + 1 = −1. Yes.
- (2) Limit exists? Polynomial → continuous everywhere → limit is −1. Yes.
- (3) Equal? −1 = −1. Continuous.
Your turn: f(x) = x³ − 2x at a = 1 — continuous?
Answer: continuous.
(1) f(1) = 1 − 2 = −1, defined. (2) It's a polynomial, so the limit is −1. (3) −1 = −1. All three hold.
Example 2 — removable hole: g(x) = (x²−4)/(x−2) at a = 2
- (1) Defined? g(2) = 0/0 — undefined. Already discontinuous.
- (2) Limit exists? Factor: (x−2)(x+2)/(x−2) = x+2 → 4. Yes.
- Diagnosis: fails (1) only — a removable discontinuity (hole at (2, 4)). Defining g(2) = 4 would repair it.
Your turn: g(x) = (x²−9)/(x+3) at a = −3 — diagnose.
Answer: fails (1) only — removable.
g(−3) is 0/0, undefined. But (x−3)(x+3)/(x+3) = x−3 → −6, so the limit exists. Defining g(−3) = −6 repairs it.
Example 3 — jump: h(x) = |x|/x at a = 0
- (1) Defined? h(0) = 0/0 — undefined. Discontinuous already.
- (2) Limit exists? Left: x < 0 gives −x/x = −1; right: x/x = 1. Disagree → no two-sided limit.
- Diagnosis: fails (1) and (2) — a jump discontinuity. No single-point redefinition can fix a jump.
Your turn: h(x) = {−1, x < 0; 1, x ≥ 0} at a = 0 — diagnose.
Answer: fails (2) — jump.
h(0) = 1 exists, but the left limit (−1) and right limit (1) disagree, so no two-sided limit exists. No single-point redefinition can fix a jump.
Example 4 — the trap: piecewise f(x) = {2x, x < 2; x+3, x ≥ 2} at a = 2
- (1) Defined? f(2) = 2+3 = 5. Yes — don’t stop here!
- (2) Limit exists? Left: 2x → 4. Right: x+3 → 5. 4 ≠ 5 → no two-sided limit.
- Diagnosis: fails (2) — jump discontinuity, despite f(2) existing.
Your turn: f(x) = {x², x < 1; 2x, x ≥ 1} at a = 1 — continuous?
Answer: no — jump at 1.
f(1) = 2 exists, but the left limit is 1² = 1 while the right limit is 2·1 = 2. Fails condition (2) despite f(1) existing.
Memorization tips
- One counterexample per condition: (x²−1)/(x−1) at 1 kills (1); ⌊x⌋ at 1 kills (2); the 5-at-0 function kills (3). Memorize the trio.
- Order matters: 1, 2, 3. Defined? Limit exists? Equal? The fastest failure mode is checked first.
- “Plug in” = condition (3): whenever you substitute directly, you are asserting continuity. Ask yourself if you’ve earned it.
- Polynomials: always safe. Sums, products, powers of x preserve continuity — direct substitution never fails for polynomials.
- Piecewise: check the seams. The formulas are usually continuous; the joints are where jumps hide. Always compute both one-sided limits at a seam.
- Name the zoo: removable (fixable by one point), jump (sides disagree), infinite (asymptote), oscillating. Naming is diagnosing.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and this law is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What are the three conditions for continuity at a?
(1) f(a) is defined, (2) lim(x→a) f(x) exists, (3) the limit equals f(a). All three must hold — each has its own counterexample.
Why isn't 'f(a) exists' enough for continuity?
It's only condition (1). The piecewise function {2x, x<2; x+3, x≥2} has f(2) = 5 but jumps (left limit 4 ≠ right limit 5) — failing condition (2).
What is a removable discontinuity?
The limit exists but f(a) is missing or at the wrong height — e.g. (x²−1)/(x−1) at x = 1. It's 'removable' because redefining a single point repairs it.
Why are polynomials continuous everywhere?
Continuity is preserved by sums, products, powers, and constant multiples (the limit laws), and x and constants are continuous. Polynomials are built from exactly those operations.
What's the difference between a jump and a removable discontinuity?
Removable: the two-sided limit exists (fixable by one point). Jump: the one-sided limits exist but disagree (like |x|/x at 0) — no single-point fix is possible.
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