Definition D1
lim(x→c) 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 − c| < δ.
Proof.
(1)The condition 0 < |x − c| excludes x = c. f(c) does not enter this definition.convention
(2)f need not have a value at c. The limit is determined only by the surroundings of c.convention
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Definition D2
f is continuous at c means that all three of the following hold. (i) f(c) is defined. (ii) lim(x→c) f(x) exists. (iii) That limit equals f(c).
Proof.
(1)(ii) and (iii) have meaning only through D1. The definition of continuity is laid on top of the definition of a limit.D1
(2)(i) is a condition on the domain of f and uses no limit.convention
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Theorem T1
f is continuous at c if and only if the following holds. For every ε > 0 there exists δ > 0 such that |f(x) − f(c)| < ε for every x with |x − c| < δ.
Proof.
(1)Suppose f is continuous at c. Then f(c) is defined and lim(x→c) f(x) = f(c).assumption, D2
(2)Suppose ε > 0 is given. There is δ > 0 such that |f(x) − f(c)| < ε whenever 0 < |x − c| < δ.(1), D1
(3)Take x with |x − c| < δ. Since 0 ≤ |x − c|, either |x − c| = 0 or 0 < |x − c|.A1, A14
(4)If |x − c| = 0, then x = c and |f(c) − f(c)| = |0| = 0 < ε.(3), assumption, A2, A6
(5)If 0 < |x − c|, then |f(x) − f(c)| < ε by (2).(2), (3)
(6)Therefore |f(x) − f(c)| < ε for every x with |x − c| < δ.(3), (4), (5)
(7)Conversely, suppose that condition holds. Since f(c) appears in the formula, f(c) is defined.assumption
(8)If 0 < |x − c| < δ, then |x − c| < δ, so |f(x) − f(c)| < ε.(7)
(9)Therefore lim(x→c) f(x) = f(c), all three conditions hold, and f is continuous at c.(7), (8), D1, D2
(10)The two conditions differ only at x = c. At that x the inequality becomes 0 < ε, which always holds, so it may be excluded or included.(4)
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Theorem T2
Even if the limit exists, without a function value there is no continuity.
Proof.
(1)Take the function with domain all reals x ≠ 0 and f(x) = x, and c = 0.convention
(2)For ε > 0, let δ = ε; then for x with 0 < |x − 0| < δ, |f(x) − 0| = |x| < ε.(1), A13
(3)Therefore lim(x→0) f(x) = 0 exists.(2), D1
(4)0 is not in the domain, so f(0) is not defined.(1)
(5)(i) fails, so f is not continuous at 0.(4), D2
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Theorem T3
Even if the function value exists, without a limit there is no continuity.
Proof.
(1)Take the function with f(x) = 1 for x > 0 and f(x) = −1 for x ≤ 0, and c = 0.convention
(2)f(0) = −1, so (i) holds.(1)
(3)Suppose lim(x→0) f(x) = L.assumption
(4)For ε = 1 there is δ > 0 such that |f(x) − L| < 1 whenever 0 < |x| < δ.(3), A12, D1
(5)0 < δ/2 < δ and |δ/2| = δ/2, so x = δ/2 satisfies that condition.(4), A10, A3
(6)δ/2 > 0, so f(δ/2) = 1 and |1 − L| < 1.(1), (4), (5)
(7)At x = −δ/2, |−δ/2| = δ/2, so 0 < |x| < δ.(5), A4, A3
(8)−δ/2 < 0, so f(−δ/2) = −1 and |−1 − L| < 1.(1), (4), (7), A10, A13
(9)|−1 − L| = |−(1 + L)| = |1 + L| = |L − (−1)|A4, A13
(10)(1 − L) + (L − (−1)) = 1 − (−1) = 1 + 1 = 2A11, A13, A15
(11)|2| ≤ |1 − L| + |L − (−1)|(10), A5
(12)|1 − L| + |L − (−1)| < 1 + |L − (−1)| < 1 + 1 = 2(6), (8), (9), A9, A7, A15
(13)0 < 2, so |2| = 2, giving 2 < 2, a contradiction.(11), (12), A12, A3, A7
(14)Therefore the limit does not exist, (ii) fails, and f is not continuous at 0.(13), A8, D2
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Theorem T4
Even if the function value and the limit both exist, if they differ there is no continuity.
Proof.
(1)Take the function with g(x) = x for x ≠ 0 and g(0) = 1, and c = 0.convention
(2)g(0) = 1, so (i) holds.(1)
(3)For ε > 0, let δ = ε; then x with 0 < |x| < δ has x ≠ 0, so g(x) = x and |g(x) − 0| = |x| < ε.(1), A2, A13
(4)Therefore lim(x→0) g(x) = 0 and (ii) holds.(3), D1
(5)If 0 = 1, then 0 < 1 gives 0 < 0, a contradiction. Therefore 0 ≠ 1.A12, A8
(6)The limit 0 and the value g(0) = 1 differ, so (iii) fails.(2), (4), (5)
(7)g is not continuous at 0.(6), D2
(8)Changing the value of g(0) from 1 to 0 makes all three conditions hold. A failure that can be removed by changing a single function value is called a removable discontinuity.(4), (7), D2
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