Definition D1
lim(x→c) f(x) = L means the following. For every ε > 0 there is a δ > 0 such that |f(x) − L| < ε for every x with 0 < |x − c| < δ.
Proof.
(1)ε is given first, and δ is chosen afterward. δ may depend on ε.convention
(2)Swapping the order of the two quantifiers gives a different meaning. The definition includes this order.convention
(3)The condition 0 < |x − c| excludes x = c. The value f(c) does not enter this definition.convention
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Lemma L1
If δ > 0, then there exists an x satisfying 0 < |x − c| < δ.
Proof.
(1)Set x = c + δ/2.convention
(2)x − c = δ/2(1), A13
(3)0 < δ/2assumption, A6
(4)|x − c| = |δ/2| = δ/2(2), (3), A3
(5)δ/2 + δ/2 = δA7
(6)δ/2 < δ/2 + δ/2 = δ(3), (5), A8
(7)0 < |x − c| < δ(3), (4), (6)
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Theorem T
If lim(x→c) f(x) = L₁ and lim(x→c) f(x) = L₂, then L₁ = L₂.
Proof.
(1)Suppose L₁ ≠ L₂.assumption
(2)L₁ − L₂ ≠ 0, so |L₁ − L₂| ≠ 0 and 0 < |L₁ − L₂|.(1), A5, A4, A3
(3)Set ε = |L₁ − L₂|/2. Then 0 < ε.(2), A6
(4)For this ε there is a δ₁ > 0 such that 0 < |x−c| < δ₁ implies |f(x) − L₁| < ε.assumption, (3), D1
(5)For the same ε there is a δ₂ > 0 such that 0 < |x−c| < δ₂ implies |f(x) − L₂| < ε.assumption, (3), D1
(6)Set δ = min(δ₁, δ₂). δ is δ₁ or δ₂, so 0 < δ.(4), (5), A15
(7)Pick an x with 0 < |x − c| < δ.(6), L1
(8)δ ≤ δ₁ and δ ≤ δ₂, so that x satisfies the conditions of both (4) and (5).(6), (7), A15, A9
(9)|f(x) − L₁| < ε and |f(x) − L₂| < ε.(4), (5), (8)
(10)|L₁ − f(x)| = |−(f(x) − L₁)| = |f(x) − L₁|A14, A2
(11)(L₁ − f(x)) + (f(x) − L₂) = L₁ − L₂A12
(12)|L₁ − L₂| ≤ |L₁ − f(x)| + |f(x) − L₂|(11), A1
(13)|L₁ − f(x)| + |f(x) − L₂| < ε + ε(9), (10), A11
(14)ε + ε = |L₁ − L₂|(3), A7
(15)|L₁ − L₂| < |L₁ − L₂|(12), (13), (14), A9
(16)This does not hold. Therefore L₁ = L₂.(15), A10
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Corollary C
The symbol lim(x→c) f(x) refers to a single value.
Proof.
(1)There cannot be two L satisfying D1.T
(2)If such an L exists, it is unique, so the name lim(x→c) f(x) can be attached to it.(1), D1
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