158 Solutions and Hints for Selected Exercises (b) The limit is calculated as follows: lim n→∞ 3 pn3 +3n2 −n =lim n→∞ 3 √n3 +3n2 −n 3 p(n3 +3n2)2 +n 3 √n3 +3n2 +n2 3 p(n3 +3n2)2 +n 3 √n3 +3n2 +n2) =lim n→∞ 3n2 3 p(n3 +3n2)2 +n 3 √n3 +3n2 +n2 =lim n→∞ 3n2 3 pn6(1+3/n)2 +n 3 pn3(1+3/n)+n2 =lim n→∞ 3n2 n2 3 p(1+3/n)2 + 3 p(1+3/n)+1 =lim n→∞ 3 3 p(1+3/n)2 + 3 p(1+3/n)+1 =1. (c) We use the result in par (a) and part (b) to obtain lim n→∞ ( 3 pn3 +3n2 −pn2 +1) =lim n→∞ 3 pn3 +3n2 −n+n−pn2 +1 =lim n→∞ 3 pn3 +3n2 −n +lim n→∞ n−pn2 +1 =1−1/2=1/2. Using a similar technique, we can find the following limit: lim n→∞ 3 pan3 +bn2 +cn+d−pαn2 +βn+γ , where a>0 and α>0. SECTION 2.3 Exercise 2.3.1. (a) Clearly, a1 <2. Suppose that ak <2 for k ∈N. Then ak+1 =p2+ak <√2+2=2. By induction, an <2 for all n∈N. (b) Clearly, a1 =√2<p2+√2=a2. Suppose that ak <ak+1 for k ∈N. Then ak +2<ak+1 +2, which implies p ak +2<pak+1 +2. Thus, ak+1 <ak+2. By induction, an <an+1 for all n∈N. Therefore, {an}is an increasing sequence. (c) By the monotone convergence theorem, limn→∞an exists. Let ℓ =limn→∞an. Since an+1 = √2+an and limn →∞an+1 =ℓ, we have ℓ =√2+ℓ or ℓ2 =2+ℓ.
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