Definition D1
The circle of radius 1 centered at the origin O is called the unit circle, and A = (1, 0). Let P = (cos x, sin x) be the point where the ray making angle x counterclockwise from the positive x-axis meets the unit circle, and let the sector S be the region bounded by the two radii OA, OP and the arc. Define tan x = sin x / cos x.
Proof.
(1)The angle x is measured in radians. The central angle of the sector S is x and its radius is 1.convention
(2)If 0 < x < π/2, then 0 < cos x, so tan x has a value.convention, A5
(3)A point on the ray OP is (t cos x, t sin x) for t > 0, and at t = 1/cos x its x-coordinate becomes 1. Let Q be its intersection with the tangent line drawn at A; then Q = (1, tan x).convention, A5, A12, A20
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Definition D2
lim(x→0) f(x) = L means the following. For every ε > 0 there exists δ > 0 such that |f(x) − L| < ε for every x with 0 < |x| < δ.
Proof.
(1)ε is given first, and δ is chosen afterward.convention
(2)The condition 0 < |x| excludes x = 0. sin x / x has no value at x = 0, so this exclusion is necessary.convention
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Lemma L1
If 0 < x < π/2, then sin x ≤ x ≤ tan x.
Proof.
(1)[△OAP] = ½ × 1 × sin x = sin x / 2. The base is OA and the height is the y-coordinate of P.D1, A2, A20
(2)[S] = ½ × 1² × x = x / 2D1, A3, A20
(3)[△OAQ] = ½ × 1 × tan x = tan x / 2. The base is OA and the height is the y-coordinate of Q.D1, A2, A20
(4)△OAP ⊆ S ⊆ △OAQA4
(5)sin x / 2 ≤ x / 2 ≤ tan x / 2(1), (2), (3), (4), A1
(6)Multiplying both sides by 2 gives sin x ≤ x ≤ tan x.(5), A16, A10, E, A20
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Lemma L2
If 0 < |x| < π/2, then cos x ≤ sin x / x ≤ 1.
Proof.
(1)First suppose 0 < x < π/2.assumption
(2)sin x ≤ x ≤ tan x(1), L1
(3)0 < 1/x.(1), A12
(4)Multiplying both sides of sin x ≤ x by 1/x gives sin x / x ≤ 1.(2), (3), A10, A20
(5)0 < cos x, so multiplying both sides of x ≤ tan x = sin x / cos x by cos x gives x cos x ≤ sin x.(1), (2), D1, A5, A10, A20
(6)Multiplying again by 1/x gives cos x ≤ sin x / x.(3), (5), A10, A20
(7)Therefore cos x ≤ sin x / x ≤ 1 on 0 < x < π/2.(4), (6)
(8)Now suppose −π/2 < x < 0 and let u = −x. Then 0 < u < π/2.assumption, A20
(9)sin x / x = (−sin u) / (−u) = sin u / u(8), A6, A20
(10)cos x = cos(−u) = cos u(8), A7
(11)Applying (7) to u gives cos u ≤ sin u / u ≤ 1, so cos x ≤ sin x / x ≤ 1.(7), (9), (10)
(12)If 0 < |x| < π/2, then x ≠ 0, so x > 0 or x < 0, and these two cases are all.(7), (11), A17, A21
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Lemma L3
lim(x→0) cos x = 1
Proof.
(1)Suppose 0 < x < π/2. Then 0 < x/2 < x < π/2.assumption, A14, A11
(2)0 < sin(x/2) and sin(x/2) ≤ x/2.(1), A5, L1
(3)sin²(x/2) ≤ (x/2)²(2), A13
(4)1 − cos x = 2 sin²(x/2) ≤ 2 (x/2)² = x²/2(3), A9, A16, A10, E, A20
(5)cos x ≤ 1, so 0 ≤ 1 − cos x.A8, A20
(6)If −π/2 < x < 0, then cos x = cos(−x) and (−x)² = x², so the same inequality holds.(4), A7, A20
(7)Therefore, if 0 < |x| < π/2, then |cos x − 1| = 1 − cos x ≤ x²/2 = |x|²/2.(4), (5), (6), A17, A21, A20
(8)Suppose ε > 0 is given. Let δ = min(ε, 1, π/2).assumption
(9)δ is one of ε, 1, π/2, and all three are positive, so 0 < δ.(8), A15, A16
(10)If 0 < |x| < δ, then δ ≤ 1, so |x| < 1, and multiplying both sides by |x| gives |x|² ≤ |x|.(8), A15, A11, A10, A20
(11)0 < |x|², so |x|²/2 < |x|² ≤ |x| < δ ≤ ε.(8), (10), A10, A14, A15, A11
(12)Therefore |cos x − 1| < ε, and lim(x→0) cos x = 1.(7), (11), A11, D2
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Theorem T
lim(x→0) sin x / x = 1
Proof.
(1)cos x ≤ sin x / x ≤ 1 on 0 < |x| < π/2.L2
(2)lim(x→0) cos x = 1L3
(3)lim(x→0) 1 = 1A19
(4)Both sides go to 1 and sin x / x lies between them, so lim(x→0) sin x / x = 1.(1), (2), (3), A18, D2
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