Calculus I › Limits › full formula sheet
Say it: “the limit as x approaches a of f of x over g of x equals L over M, provided M is not zero”
The quotient law for limits
Split a limit across a fraction — unless the denominator is heading for zero.
Notation on this page: L = limx→a f(x) and M = limx→a g(x) both exist (finite), and M ≠ 0. That last condition is the whole story.
Before this lesson: The product law
Notation in this lesson
- L/M
- the top limit over the bottom limit
- M ≠ 0
- the denominator's limit can't be zero — the law's one extra demand
- 0/0
- not an answer — a signal to simplify first
Where it comes from
The quotient law is the product law wearing a disguise: f/g is f · (1/g). Its one extra demand — the denominator's limit cannot be zero — is where all the drama lives. Two ways the naive “split and plug” dies:
The intuition: if g(x) hugs M ≠ 0, then g stays safely away from zero near a, so 1/g(x) hugs 1/M without blowing up. Then f/g = f · (1/g) is just the product law. Everything hard is proving that “safely away from zero” part — which is exactly what the derivation does.
Derivation
We prove the reciprocal law first (lim 1/g = 1/M), then the quotient law follows from the product law. Assume lim(x→a) g(x) = M ≠ 0.
Key steps shown; the argument above is complete. Notice the architecture: reciprocal law (needs M ≠ 0) + product law = quotient law. Laws compose — that is why the limit laws are taught as a family.
How to use it
The procedure, every time:
- Check the denominator’s limit first. If lim g = M ≠ 0: split and divide — you are done. If lim g = 0: stop — the law is off; go to step 3.
- Split: evaluate lim f and lim g separately, then compute L/M.
- Denominator → 0? Analyze directly: factor and cancel (0/0 signal), rationalize, or do a one-sided sign analysis (nonzero numerator over vanishing denominator → infinite).
- Read 0/0 as an instruction, not an answer: it means “simplify me first.”
The 0/0 playbook
When splitting gives 0/0, try in order: factor and cancel (polynomials), rationalize (roots), key trig limits (sin x / x). Then re-evaluate — the new pieces usually converge.
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→2) (x²+1)/(x+3)
- Check the denominator. lim(x→2) (x+3) = 5 ≠ 0 — safe to split.
- Evaluate each piece. Numerator → 5, denominator → 5.
- Divide. 5/5 = 1.
Your turn: limx→1 (x²+2)/(x+1) = ?
Answer: 3/2.
Denominator → 2 ≠ 0, so split: (1+2)/(1+1) = 3/2.
Example 2 — the 0/0 signal: lim(x→1) (x²−1)/(x−1)
- Try the law: numerator → 0, denominator → 0. Stop — simplify first.
- Factor and cancel. (x−1)(x+1)/(x−1) = x+1 for x ≠ 1. (Why legal? The limit only cares about x near 1, never at 1.)
- Re-evaluate: lim(x→1) (x+1) = 2.
Your turn: limx→3 (x²−9)/(x−3) = ?
Answer: 6.
0/0 — factor: (x−3)(x+3)/(x−3) = x+3 → 6.
Example 3 — another 0/0: lim(x→5) (x²−25)/(x−5)
- 0/0 signal — factor: (x−5)(x+5)/(x−5) = x+5 (x ≠ 5).
- Re-evaluate: 5 + 5 = 10.
- Pattern check: lim(x→a) (x²−a²)/(x−a) = 2a always — this is secretly the derivative of x² at a.
Your turn: limx→2 (x³−8)/(x−2) = ?
Answer: 12.
0/0 — factor: (x−2)(x²+2x+4)/(x−2) = x²+2x+4 → 4+4+4 = 12.
Example 4 — the trap: lim(x→0) (x+1)/x
- Check the denominator. lim(x→0) x = 0 — the law is off.
- One-sided analysis: numerator → 1 (nonzero), denominator → 0. From the right: 1/0+ = +infinity; from the left: 1/0− = −infinity.
- Verdict: the two-sided limit DNE (the one-sided infinite limits disagree).
Your turn: limx→0 (x+2)/x = ?
Answer: DNE (two-sided).
Right side: 2/0+ = +∞; left side: 2/0− = −∞. The sides disagree.
Memorization tips
- Denominator first: the first thing you check in any quotient limit is where the denominator is heading. Nonzero → split. Zero → simplify or go one-sided.
- 0/0 is an instruction, not an answer: it means “factor me, rationalize me, or trig-limit me.” The same signal hides 2, 10, and 1/2.
- f/g = f · (1/g): the quotient law is the product law plus the reciprocal law. If you forget the statement, rebuild it from this.
- The reciprocal proof’s key image: g → M ≠ 0 means g eventually stays more than |M|/2 away from zero — that “buffer zone” is what keeps 1/g under control.
- Canceling is legal because limits ignore the point itself: (x−1)/(x−1) = 1 for x ≠ 1 is all the limit needs.
- After canceling, you usually land in direct substitution: the simplified expression is continuous at the point — that is why the 0/0 playbook ends with “just plug in.”
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 quotient law for limits?
If lim(x→a) f(x) = L and lim(x→a) g(x) = M exist, are finite, and M ≠ 0, then lim(x→a) f(x)/g(x) = L/M: the limit of a quotient is the quotient of the limits.
Why can't the denominator's limit be zero?
The proof needs 1/g to stay bounded, which requires g to stay away from zero — impossible if g → 0. When the denominator → 0, analyze directly: factor, rationalize, or check one-sided infinite behavior.
What does 0/0 mean when I split a limit?
It's a signal to simplify first, not an answer. Factor and cancel, rationalize, or use trig limits, then re-evaluate — the same 0/0 signal can hide 2, 10, or 1/2.
How does the quotient law follow from the product law?
Write f/g as f·(1/g). The reciprocal law gives lim(1/g) = 1/M (this is where M ≠ 0 is used), and the product law multiplies: L·(1/M) = L/M.
Is it legal to cancel (x−1) from (x²−1)/(x−1)?
Yes — for x ≠ 1 the expressions are identical, and a limit only cares about x near 1, never at 1. After canceling, x+1 is continuous, so the limit is 2.
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