69 Therefore, lim x→−1 x2 +6x+5 x+1 = lim x→−1 x+5=4. Theorem 3.2.2 (Cauchy’s criterion) Let f : D→Rand let x0 be a limit point of D. The following are equivalent: (i) f has a limit at x0. (ii) For anyε >0, there exists δ >0 such that | f (r)−f (s)| <ε whenever r,s ∈Dand 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 | f (x)−ℓ| < ε 2 whenever x ∈Dand 0<|x−x0| <δ. Thus, for r,s ∈Dwith 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 Dsuch that limn→∞un =x0 and un̸ =x0 for every n. We are assuming that givenε >0, there exists δ >0 such that | f (r)−f (s)| <ε whenever r,s ∈Dand 0<|r−x0| <δ,0<|s−x0| <δ. Since limn→∞un =x0, there exists N∈Nsatisfying 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 ∈Rsuch 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 Dsuch that limn→∞xn =x0 and xn̸ =x0 for every n. By the previous argument, there exists ℓ′ ∈Rsuch that lim n→∞ f (xn) =ℓ′. Fix anyε >0 and let δ >0 satisfy (3.1). There exists K∈Nsuch that |un −x0| <δ and|xn −x0| <δ for all n≥K. Then| f (un)−f (xn)| <ε for suchn. Lettingn→∞, 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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