37 Proof: Suppose {an}converges toaandb. Then givenε >0, there exist positive integers N1 and N2 such that |an −a| <ε/2 for all n≥N1 and |an −b| <ε/2 for all n≥N2. Let N=max{N1,N2}. Then |a−b|≤|a−aN|+|aN−b| <ε/2+ε/2=ε. Since ε >0 is arbitrary, it follows that |a−b| =0 (see Exercise 1.4.10) and hence a=b. The next result shows that (non-strict) inequalities are preserved “in the limit”. Theorem 2.1.2 — Comparison Theorem. Let {an} and {bn} be sequences of real numbers. Suppose that: (i) limn →∞ an =a and limn →∞ bn =bfor some a,b∈R, (ii) an ≤bn for all n∈N. Then a≤b. Proof: For anyε >0, there exist N1,N2 ∈Nsuch that a− ε 2 <an <a+ ε 2 , for n≥N1, b− ε 2 <bn <b+ ε 2 , for n≥N2. Choose N=max{N1,N2}. Then a− ε 2 <aN ≤bN <b+ ε 2 . Thus, a<b+ε for anyε >0. From this it follows that a≤b (see Exercise 1.4.10). □ Theorem 2.1.3 — Squeeze Theorem. Let {an}, {bn}, and {cn} be sequences of real numbers. Suppose that: (i) an ≤bn ≤cn for all n∈N, (ii) limn →∞ an =ℓ =limn →∞ cn. Then limn →∞ bn =ℓ. Proof: Fix any ε >0. Since limn →∞ an =ℓ, there exists N1 ∈Nsuch that ℓ−ε <an <ℓ+ε for all n≥N1. Similarly, since limn →∞ cn =ℓ, there exists N2 ∈Nsuch that ℓ−ε <cn <ℓ+ε for all n≥N2. Let N=max{N1,N2}. Then, for n≥N, we have ℓ−ε <an ≤bn ≤cn <ℓ+ε, which implies |bn −ℓ| <ε. Therefore, limn →∞ bn =ℓ. □
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