158 Solutions and Hints for Selected Exercises (b) The limit is calculated as follows: √ p √ 3 3 3 2 p n3 + 3n2 − n (n3 + 3n2)2 + n n3 + 3n2 + n lim 3 n3 + 3n2 − n = lim p √ 3 3 n→∞ n→∞ (n3 + 3n2)2 + n n3 + 3n2 + n2) 3n2 = lim p √ n→∞ 3 (n3 + 3n2)2 + n 3 n3 + 3n2 + n2 3n2 = lim p p n→∞ 3 n6(1 + 3/n)2 + n 3 n3(1 + 3/n)+ n2 p 3n2 p = lim n→∞ n2 3 (1 + 3/n)2 + 3 (1 + 3/n)+ 1 = lim p 3 = 1. p n→∞ 3 (1 + 3/n)2 + 3 (1 + 3/n)+ 1 (c) We use the result in par (a) and part (b) to obtain p p p p n3 + 3n2 − n2 + 1)= n3 + 3n2 − n + n− n2 + 1 lim ( 3 lim 3 n→∞ n→∞ p p = lim 3 n3 + 3n2 − n + lim n − n2 + 1 = 1 − 1/2 = 1/2. n→∞ n→∞ Using a similar technique, we can fnd the following limit: p p lim 3 an3 + bn2 + cn + d − αn2 + β n + γ , n→∞ 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 p √ ak+1 = 2+ ak < 2+ 2 = 2. By induction, an < 2 for all n ∈ N. √ p √ (b) Clearly, a1 = 2 < 2 + 2 = a2. Suppose that ak < ak+1 for k ∈ N. Then ak + 2 < ak+1 + 2, which implies p p ak + 2 < ak+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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