37 Proof: Suppose {an} converges to a and b. Then given ε > 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 = b for some a,b ∈ R, (ii) an ≤ bn for all n ∈ N. Then a ≤ b. Proof: For any ε > 0, there exist N1,N2 ∈ N such that ε ε a− < an < a + , for n ≥ N1, 2 2 ε ε b− < bn < b + , for n ≥ N2. 2 2 Choose N = max{N1,N2}. Then ε ε a− < aN ≤ bN < b + . 2 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 any ε > 0. Since limn →∞ an = ℓ, there exists N1 ∈ N such that ℓ − ε < an <ℓ + ε for all n ≥ N1. Similarly, since limn →∞ cn = ℓ, there exists N2 ∈ N such 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