Calculus I › Limits › full formula sheet
The key trig limit: sin x / x
The 0/0 limit that unlocks every trig derivative — in radians only.
On this page x is in radians — the theorem is false in degrees. Every “matching arguments” move below means rewriting so the sine’s input equals the denominator.
Before this lesson: The squeeze theorem
Notation in this lesson
- x
- in radians — always
- sin x / x
- a 0/0 form at x = 0
- π/180
- the value you get in degree mode — not 1
Where it comes from
Direct substitution gives 0/0 — indeterminate, a signal to work, not an answer. This limit matters because it is the key step in proving d/dx[sin x] = cos x: the difference quotient for sine collapses to exactly (sin h)/h pieces. No squeeze, no trig derivatives.
The naive traps:
The intuition: for tiny x, sin x ≈ x (the sine graph hugs the line y = x near 0), so their ratio ≈ 1. The derivation turns “hugs” into the sandwich cos x < (sin x)/x < 1.
Derivation
The classic geometric squeeze. For 0 < x < π/2, compare three areas in the unit circle: triangle OAB, sector OAB, triangle OAB′ (with B′ = (1, tan x)).
Key steps shown; the argument above is complete. Note what made it work: the sector area formula ½r²θ requires radian measure — that is the deep reason the theorem is radian-specific. In degrees the sector area is (π/360)·x, and the π/180 leaks into the answer.
How to use it
The procedure, every time:
- Confirm radians. Degree-mode inputs need converting first (or accept the π/180 factor).
- Match the arguments: rewrite so the sine’s input equals the denominator: sin(5x)/x = 5 · sin(5x)/(5x).
- Apply: each matched (sin □)/□ → 1; multiply by whatever constants the matching introduced.
- Combine with product/quotient laws for ratios of trig expressions.
The matching-arguments move
lim(x→0) sin(7x)/(2x): the sine says 7x but the denominator says 2x. Insert the match: (7/2) · sin(7x)/(7x) → (7/2)·1 = 7/2. The constant you introduce (7/2) is the whole answer — the trig part always contributes exactly 1.
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→0) sin(5x)/x
- Match arguments: sin(5x)/x = 5 · sin(5x)/(5x).
- Apply the theorem: sin(5x)/(5x) → 1 (as x → 0, so does 5x → 0).
- Result: 5 · 1 = 5.
Your turn: limx→0 sin(7x)/x = ?
Answer: 7.
sin(7x)/x = 7·sin(7x)/(7x) → 7·1 = 7.
Example 2 — ratio of sines: lim(x→0) sin(3x)/sin(2x)
- Match both: [3 · sin(3x)/(3x)] / [2 · sin(2x)/(2x)].
- Each trig fraction → 1: (3·1)/(2·1) = 3/2.
Your turn: limx→0 sin(4x)/sin(5x) = ?
Answer: 4/5.
[4·sin(4x)/(4x)] / [5·sin(5x)/(5x)] → (4·1)/(5·1) = 4/5.
Example 3 — reciprocal: lim(x→0) x/sin(4x)
- Flip it: x/sin(4x) = (1/4) · (4x)/sin(4x) = (1/4) · 1/[(sin 4x)/(4x)].
- The denominator → 1 by the theorem, so the fraction → (1/4) · 1 = 1/4.
Your turn: limx→0 x/sin(6x) = ?
Answer: 1/6.
x/sin(6x) = (1/6)·(6x)/sin(6x) = (1/6)·1/[(sin 6x)/(6x)] → 1/6.
Example 4 — the trap: lim(x→0) sin(x°)/x (degrees!)
- Convert: sin(x°) = sin(πx/180). Let u = πx/180 (u → 0).
- Rewrite: sin(u)/(180u/π) = (π/180) · sin(u)/u.
- Apply: (π/180) · 1 = π/180 ≈ 0.01745 — not 1!
Your turn: limx→0 sin(2x°)/x = ? (degrees)
Answer: π/90 ≈ 0.0349.
sin(2x°) = sin(πx/90); with u = πx/90, this is (π/90)·sin(u)/u → π/90. Radians matter!
Memorization tips
- Radians or it didn’t happen: the theorem is radian-specific. Degree inputs pick up a π/180 factor.
- Match the arguments: sin(5x)/x = 5·sin(5x)/(5x). The introduced constant is the answer; the trig part is always 1.
- The coefficient is the answer: lim sin(ax)/x = a, lim sin(ax)/sin(bx) = a/b, lim x/sin(ax) = 1/a. One pattern, three faces.
- Why it matters: this limit is the key step in d/dx[sin x] = cos x. It’s not trivia — it’s load-bearing.
- Even function, free left limit: sin x/x is even, so proving x → 0+ gives both sides.
- The squeeze bread: cos x < (sin x)/x < 1. Picture the sector sandwiched between the two triangles — that image is the proof.
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 lim(x→0) sin x / x?
1 — provided x is in radians. It's a 0/0 indeterminate form resolved by a geometric squeeze: cos x < (sin x)/x < 1 near 0.
Why must x be in radians?
The proof uses the sector area formula ½r²θ, which requires radian measure. In degrees the limit is π/180 ≈ 0.01745, not 1.
How do I compute lim(x→0) sin(5x)/x?
Match the arguments: sin(5x)/x = 5·sin(5x)/(5x) → 5·1 = 5. The introduced constant is the answer; the trig part always contributes 1.
Why is this limit so important?
It's the key step in proving d/dx[sin x] = cos x — the difference quotient for sine reduces to (sin h)/h pieces. Without it, no trig derivatives.
Can I just cancel sin(3x)/sin(2x) to 3x/2x?
The value 3/2 is correct, but 'canceling the sines' is illegal. The legal route is matching arguments: [3·sin(3x)/(3x)]/[2·sin(2x)/(2x)] → 3/2.
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