43 Now let ε > 0. Since limn →∞ bn = b, there exists N2 ∈ N such that b2ε |bn − b| < for all n ≥ N2. 2 Let N = max{N1,N2}. By (2.2), one has 1 1 2|bn − b| − ≤ < ε for all n ≥ N. b2 bn b 1 1 It follows that limn →∞ = . bn b Finally, we can apply part (iii) and have an 1 a lim = lim an = . n→∞ bn n→∞ bn b The proof is now complete. □ ■ Example 2.2.1 Consider the sequence {an} given by 3n2 − 2n + 5 an = (2.3) 1 − 4n + 7n2 . Dividing numerator and denominator by n2, we can write 3 − 2/n+ 5/n2 an = (2.4) 1/n2 − 4/n + 7 Therefore, by the limit theorems above, 3− 2/n + 5/n2 limn→∞ 3 − limn →∞ 2/n + limn →∞ 5/n2 3 lim an = lim = = . (2.5) n→∞ n→∞ 1/n2 − 4/n + 7 limn→∞ 1/n2 − limn→∞ 4/n + limn →∞ 7 7 √ ■ Example 2.2.2 Let an = n b, where b > 0. Consider the case where b > 1. In this case, an > 1 for every n. By the binomial theorem, n b = a =(an − 1 + 1) n ≥ 1 + n(an − 1). n This implies b − 1 0 < an − 1 ≤ . n b − 1 For each ε > 0, choose N > . It follows that for n ≥ N, ε b − 1 b − 1 |an − 1| = an − 1 < ≤ < ε. n N Thus, limn →∞ an = 1. In the case where b = 1, it is obvious that an = 1 for all n and, hence, limn →∞ an = 1.
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