Calculus I › Limits › full formula sheet
The key trig limit: (1 − cos x) / x
Sin x / x's quieter partner — and the reason cosine's derivative has its minus sign.
On this page x is in radians. This limit is the key step in proving d/dx[cos x] = −sin x — the 0 here is where that minus sign’s story begins.
Before this lesson: The sin x / x limit
Notation in this lesson
- x
- in radians — the theorem fails in degrees
- (1−cos x)/x
- a 0/0 form at x = 0
- 1/2
- the value of the x² variant: limx→0 (1−cos x)/x²
Where it comes from
Another 0/0 indeterminate — and the partner the sine limit needed. When you differentiate cos x from the definition, the difference quotient splits into a (sin h)/h piece and a (1 − cos h)/h piece. The first gives 1; this page’s limit gives 0 — and together they assemble d/dx[cos x] = −sin x.
The naive traps:
The intuition: near 0, cos x ≈ 1 − x²/2 (the parabola that hugs cosine), so 1 − cos x ≈ x²/2 — much smaller than x. Dividing by x still leaves something → 0; dividing by x² leaves the 1/2. The derivation makes this precise without any series.
Derivation
The conjugate trick: multiply by (1+cos x)/(1+cos x) = 1, turning 1 − cos x into 1 − cos²x = sin²x. Full proof, one row per move:
Full proof shown. This is the same conjugate move as the root law’s proof — rationalizing a difference to expose a Pythagorean identity. One trick, many theorems.
How to use it
The procedure, every time:
- Confirm radians and the shape (1 − cos □)/□ with matching arguments and a single power of x below.
- Match arguments like the sine page: (1−cos 3x)/x = 3 · (1−cos 3x)/(3x) → 3·0 = 0.
- Check the denominator’s power: x² below means 1/2, not 0 — regroup as [(sin x)/x]² · 1/(1+cos x).
- Sign flips are harmless: (cos x − 1)/x = −(1−cos x)/x → 0 either way.
The x vs x² split
Memorize the pair together: (1−cos x)/x → 0 but (1−cos x)/x² → 1/2. The x² version keeps one extra (sin x)/x factor alive: [(sin x)/x]² · 1/(1+cos x) → 1² · 1/2. If you only remember one of them, remember the 1/2 — it’s the one exams love to test.
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) (1−cos 3x)/x
- Match arguments: (1−cos 3x)/x = 3 · (1−cos 3x)/(3x).
- Apply the theorem: (1−cos 3x)/(3x) → 0.
- Result: 3 · 0 = 0.
Your turn: limx→0 (1−cos 4x)/x = ?
Answer: 0.
Match arguments: (1−cos 4x)/x = 4·(1−cos 4x)/(4x) → 4·0 = 0.
Example 2 — the trap: lim(x→0) (1−cos x)/x² = 1/2
- Do NOT apply the basic theorem — the denominator is x², not x.
- Conjugate trick: (1−cos x)/x² = sin²x/[x²(1+cos x)] = [(sin x)/x]² · 1/(1+cos x).
- Evaluate: [(sin x)/x]² → 1² = 1; 1/(1+cos x) → 1/2. Product: 1/2.
Your turn: limx→0 (1−cos 2x)/x² = ?
Answer: 2.
Conjugate: (1−cos 2x)/x² = sin²(2x)/[x²(1+cos 2x)] = [2·(sin 2x)/(2x)]²·1/(1+cos 2x) → 4·1·(1/2) = 2.
Example 3 — sign flip: lim(x→0) (cos x − 1)/x
- Factor the minus: (cos x − 1)/x = −(1−cos x)/x.
- Apply: −1 · 0 = 0.
Your turn: limx→0 (cos 3x − 1)/x = ?
Answer: 0.
Factor the minus: −(1−cos 3x)/x = −3·(1−cos 3x)/(3x) → −3·0 = 0.
Example 4 — matching with a coefficient: lim(x→0) (1−cos 2x)/(3x)
- Match arguments: let u = 2x; then (1−cos u)/(3u/2) = (2/3) · (1−cos u)/u.
- Apply: (2/3) · 0 = 0.
Your turn: limx→0 (1−cos 5x)/(2x) = ?
Answer: 0.
(1−cos 5x)/(2x) = (5/2)·(1−cos 5x)/(5x) → (5/2)·0 = 0.
Memorization tips
- Memorize the pair: (1−cos x)/x → 0, (1−cos x)/x² → 1/2. The power below decides — count it first.
- The conjugate trick is the engine: (1−cos x)(1+cos x) = sin²x. Same rationalization as the root law’s proof.
- Near 0, 1−cos x behaves like x²/2: much smaller than x (hence → 0 over x), exactly half of x² (hence → 1/2 over x²).
- Coefficients die here: unlike the sine page, the trig part contributes 0 — so any finite matching constant still gives 0.
- Why it matters: this 0 is the key step in d/dx[cos x] = −sin x. Load-bearing, not trivia.
- Radians or it didn’t happen: same warning as the sine page — the Pythagorean identity is unit-free, but the (sin x)/x → 1 step needs radians.
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) (1 − cos x)/x?
0 (x in radians). It's a 0/0 form resolved by the conjugate trick: multiply by (1+cos x)/(1+cos x) to get sin²x/[x(1+cos x)] = [(sin x)/x]·[sin x/(1+cos x)] → 1·0 = 0.
Why is lim(x→0) (1 − cos x)/x² = 1/2 and not 0?
The x² version keeps an extra (sin x)/x factor alive: [(sin x)/x]²·1/(1+cos x) → 1·(1/2). The power of x in the denominator decides — count it first.
What is the conjugate trick?
Multiplying (1−cos x) by (1+cos x)/(1+cos x) = 1 turns 1−cos x into 1−cos²x = sin²x — the same rationalization move as the root law's proof.
Why does this limit matter?
It's the key step in proving d/dx[cos x] = −sin x: the difference quotient for cosine splits into (sin h)/h and (1−cos h)/h pieces.
Does it work in degrees?
No — the proof's (sin x)/x → 1 step needs radians. The Pythagorean identity itself is unit-free, but the trig limit it leans on isn't.
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