Calculus I › Limits › full formula sheet
Say it: “the limit as x approaches a of f of x times g of x equals L times M”
The product law for limits
The limit of a product is the product of the limits — when both pieces converge.
Notation on this page: L = limx→a f(x) and M = limx→a g(x) are assumed to exist (finite).
Before this lesson: The sum / difference law
Notation in this lesson
- L·M
- the product of the two individual limits
- hypothesis
- both factor limits must exist and be finite — or the law is silent
Where it comes from
The problem this law solves is the same divide-and-conquer as the sum law, one operation over: lim(x→4) x·√x should just be 4·2 = 8. And it is — provided both factors converge. The subtlety is that the converse fails, and failing to notice costs marks:
The intuition: if f(x) hugs L and g(x) hugs M, then f(x)·g(x) hugs L·M. The error analysis is a little richer than the sum law’s: f·g − L·M = f·(g−M) + M·(f−L). Each term pairs one factor’s error with the other factor’s size — and the proof below shows both terms vanish.
Derivation
Assume lim(x→a) f(x) = L and lim(x→a) g(x) = M, both finite. We prove lim(x→a) [f(x)·g(x)] = L·M.
Key steps shown; the argument above is complete. Notice the constant multiple law hiding inside: it is the product law with g(x) = c, whose limit is c. And the sum law’s min-delta trick appears again whenever two approximations must hold at once.
How to use it
The procedure, every time:
- Check the hypothesis. Do lim f and lim g both exist (finite)? If a factor oscillates or blows up, stop — reach for the squeeze theorem or simplify first.
- Evaluate each factor separately, then multiply the results.
- Three or more factors? Apply the law repeatedly: lim(f·g·h) = lim((f·g)·h).
- Watch for disguised products: (x−2)·(1/(x−2)) looks splittable but the second factor diverges — simplify to 1 first.
When the product law is the wrong tool
If one factor’s limit does not exist, the law is off the table — but the limit itself might still exist. The classic rescue is the squeeze theorem: a bounded oscillation times something going to 0 gives 0. Pattern to memorize: bounded × vanishing = 0.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: lim(x→4) x·√x
- Check the hypothesis. lim(x→4) x = 4 and lim(x→4) √x = 2 both exist.
- Evaluate each factor. 4 and 2.
- Multiply. 4 · 2 = 8.
Your turn: limx→9 x·√x = ?
Answer: 27.
Both factors converge: lim x = 9 and lim √x = 3, so 9·3 = 27.
Example 2 — with trigonometry: lim(x→π/2) x·sin x
- Check the hypothesis. lim(x→π/2) x = π/2; lim(x→π/2) sin x = 1 (sine is continuous). Both exist.
- Multiply. (π/2) · 1 = π/2.
- Why this is safe: sin x is not oscillating wildly here — near π/2 it hugs 1. Oscillation only breaks the law when the limit truly DNE.
Your turn: limx→π x·cos x = ?
Answer: −π.
lim x = π; lim cos x = cos π = −1 (cosine is continuous). Product: −π.
Example 3 — three factors: lim(x→1) x(x+1)(x+2)
- Apply the law twice. lim[(x(x+1))·(x+2)] = lim[x(x+1)] · lim(x+2) = [lim x · lim(x+1)] · lim(x+2).
- Evaluate each: 1 · 2 · 3 = 6.
Your turn: limx→2 x(x−1)(x+1) = ?
Answer: 6.
Apply the law twice: 2·1·3 = 6.
Example 4 — the trap: lim(x→0) x·sin(1/x)
- Check the hypothesis. lim(x→0) sin(1/x) does not exist. The product law is illegal here.
- Switch tools: squeeze. Since −1 ≤ sin(1/x) ≤ 1, multiply by |x|: −|x| ≤ x·sin(1/x) ≤ |x|.
- Outer limits match: both −|x| and |x| tend to 0, so the limit is 0.
Your turn: limx→0 x²·sin(1/x) = ?
Answer: 0.
The product law is illegal (sin(1/x) oscillates), so squeeze: −x² ≤ x²sin(1/x) ≤ x² → 0.
Memorization tips
- The law goes one way: factors converging implies the product converges — but a convergent product does not imply convergent factors. x·sin(1/x) → 0 is the counterexample to memorize.
- Bounded × vanishing = 0: the pattern behind every squeeze-theorem product. When you see oscillation times something → 0, think squeeze, not product law.
- The add-and-subtract trick reappears: fg − LM = f(g−M) + M(f−L) is the same “add zero in disguise” move as the product rule proof. One trick, two theorems.
- Boundedness is the hidden hypothesis: a convergent function is bounded near the point — that is what lets the proof control the |f|·|g−M| term.
- Constant multiple is a special case: set g(x) = c. Fewer laws to memorize when you see the nesting.
- Count your applications: n factors need n−1 uses of the law. Three factors, two splits.
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 is the product law for limits?
If lim(x→a) f(x) = L and lim(x→a) g(x) = M both exist and are finite, then lim(x→a) [f(x)·g(x)] = L·M: the limit of a product is the product of the limits.
Can a product's limit exist when a factor's limit doesn't?
Yes — lim(x→0) x·sin(1/x) = 0 even though lim(x→0) sin(1/x) DNE. The product law goes one way only; this case needs the squeeze theorem.
What is the add-and-subtract trick in the proof?
We write fg − LM = f·(g−M) + M·(f−L) (adding −fM + Mf = 0). Each group pairs one factor's error with the other's size, so both vanish in the limit.
Why does the proof need f to be bounded near a?
The error term |f|·|g−M| could blow up if |f| were unbounded. Convergent functions are bounded near the point, which keeps this term under control.
How do I handle three or more factors?
Apply the law repeatedly: lim(f·g·h) = lim((f·g)·h) = L·M·N, as long as every factor's limit exists.
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