44 2.2 Limit Theorems 1 If 0 < b < 1, let c = and defne b √ 1 xn = n c = . an Since c > 1, it has been shown that limn →∞ xn = 1. This implies 1 lim an = lim = 1. n→∞ n→∞ xn Exercises 2.2.1 Find the following limits: 3n2 − 6n + 7 (a) lim , n→∞ 4n2 − 3 3 1 + 3n − n (b) lim . n→∞ 3n3 − 2n2 + 1 2.2.2 Find the following limits: √ 3n + 1 (a) lim √ √ , n→∞ n + 3 r n 2n+ 1 (b) lim . n→∞ n 2.2.3 ▶ Find the following limits if they exist: p (a) lim ( n2 + n− n), n→∞ p (b) lim ( 3 n3 + 3n2 − n), n→∞ p p n3 + 3n2 − n2 + n), (c) lim ( 3 n→∞ √ √ (d) lim ( n+ 1 − n), n→∞ √ √ (e) lim ( n+ 1 − n)/n. n→∞ 2.2.4 Find the following limits. (a) For |r| < 1 and b ∈ R, limn →∞(b + br + br2 + ··· + brn). 2 2 2 (b) lim + ··· + . n→∞ 10 102 + 10n 2.2.5 Provide counterexamples for each of the following statements: (a) If {an} and {bn} are divergent sequences, then {an + bn} is a divergent sequence. (b) If {an} and {bn} are divergent sequences, then {anbn} is a divergent sequence. (c) If {an} and {an + bn} are divergent sequences, then {bn} is a divergent sequence.
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