43 Now let ε >0. Since limn→∞bn =b, there exists N2 ∈Nsuch that |bn −b| < b2ε 2 for all n≥N2. Let N=max{N1,N2}. By (2.2), one has 1 bn − 1 b ≤ 2|bn −b| b2 <ε for all n≥N. It follows that limn→∞ 1 bn = 1 b . Finally, we can apply part (iii) and have lim n→∞ an bn =lim n→∞ an 1 bn = a b . The proof is now complete. □ ■ Example 2.2.1 Consider the sequence {an}given by an = 3n2 −2n+5 1−4n+7n2 . (2.3) Dividing numerator and denominator byn2, we can write an = 3−2/n+5/n2 1/n2 −4/n+7 (2.4) Therefore, by the limit theorems above, lim n→∞ an =lim n→∞ 3−2/n+5/n2 1/n2 −4/n+7 = limn→∞3−limn→∞2/n+limn→∞5/n 2 limn→∞1/n2 −limn →∞4/n+limn→∞7 = 3 7 . (2.5) ■ Example 2.2.2 Let an = n √b, where b>0. Consider the case where b>1. In this case, an >1 for everyn. By the binomial theorem, b=an n = (an −1+1) n ≥1+n(an −1). This implies 0<an −1≤ b−1 n . For each ε >0, choose N> b−1 ε . It follows that for n≥N, |an −1| =an −1< b−1 n ≤ b−1 N <ε. Thus, limn→∞an =1. In the case where b=1, it is obvious that an =1 for all nand, hence, limn→∞an =1.
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