69 Therefore, 2x + 6x + 5 lim = lim x + 5 = 4. x→−1 x + 1 x→−1 Theorem 3.2.2 (Cauchy’s criterion) Let f : D → R and let x0 be a limit point of D. The following are equivalent: (i) f has a limit at x0. (ii) For any ε > 0, there exists δ > 0 such that | f (r) − f (s)| < ε whenever r,s ∈ D and 0 < |r − x0| < δ ,0 < |s − x0| < δ . (3.1) Proof: Suppose limx →x0 f (x)= ℓ. Given ε > 0, there exists δ > 0 such that ε | f (x) − ℓ| < whenever x ∈ D and 0 < |x − x0| < δ . 2 Thus, for r,s ∈ D with 0 < |r − x0| < δ and 0 < |s − x0| < δ , we have | f (r) − f (s)|≤| f (r) − ℓ| + |ℓ − f (s)| < ε. Let us prove the converse. Fix a sequence {un} in D such that limn →∞ un = x0 and un ̸ = x0 for every n. We are assuming that given ε > 0, there exists δ > 0 such that | f (r) − f (s)| < ε whenever r,s ∈ D and 0 < |r − x0| < δ ,0 < |s − x0| < δ . Since limn →∞ un = x0, there exists N ∈ N satisfying 0 < |un − x0| < δ for all n ≥ N. This implies | f (un) − f (um)| < ε for all m,n ≥ N. Thus, { f (un)} is a Cauchy sequence, and hence there exists ℓ ∈ R such that limn →∞ f (un)= ℓ. We now prove that f has limit ℓ at x0 using Theorem 3.1.2. Let {xn} be a sequence in D such that limn →∞ xn = x0 and xn ̸ = x0 for every n. By the previous argument, there exists ℓ ′ ∈ R such that lim f (xn)= ℓ ′ . n→∞ Fix any ε > 0 and let δ > 0 satisfy (3.1). There exists K ∈ N such that |un − x0| < δ and |xn − x0| < δ for all n ≥ K. Then | f (un) − f (xn)| < ε for such n. Letting n → ∞, we have |ℓ − ℓ ′| ≤ ε. Thus, ℓ = ℓ ′ since ε is arbitrary. It now follows from Theorem 3.1.2 that limx →x0 f (x)= ℓ. □ The rest of this section discusses some special limits and their properties. First we introduce the notion of left and right limit point of a set.
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