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 =aand limn→∞bn =bfor some a,b∈R, (ii) an ≤bn for all n∈N. Then a≤b. Proof: For any ε >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 any ε >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 anyε >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. □
RkJQdWJsaXNoZXIy NTc4NTAz