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Lecture39-rev

Lecture39-rev - MAT A26 Lecture 39 1 Calculating Limits of...

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M AT A26 Lecture 39 1 Calculating Limits of Sequences Theorem 1. Let a k be a sequence with the property that a k = f ( k ) for k = 1 , 2 , . . . for some function with dom f [1 , ) . Then lim x →∞ f ( x ) = L = lim k →∞ a k = L lim x →∞ | f ( x ) | = = lim k →∞ | a k | = (In fact this is true if we only have a k = f ( k ) for k k 0 and not nesseccarily a k = f ( k ) for k 1 .) 1.0.1 Example Let a k = 1 - 2 k 3 ( k + 1)( k 2 + 2) . Calculate lim k →∞ a k . SOLUTION: Set f ( x ) = 1 - 2 x 3 ( x + 1)( x 2 + 2) = - 2 x 3 + 1 x 3 + x 2 + 2 x + 2 = - 2 so by the theorem, lim k →∞ a k = - 2. 1.0.2 Example Let a k = (1 - 1 k 2 ) k 2 . Calculate the limit of this sequence. SOLUTION: Set f ( x ) = (1 - 1 x 2 ) x 2 = e x 2 ln(1 - 1 x 2 ) . Now lim x →∞ x 2 ln(1 - 1 x 2 ) = lim x →∞ ln(1 - 1 x 2 1 x 2 L’H = lim x →∞ - 2 x 3 1 - 1 x 2 - 2 x 3 = lim x →∞ 1 1 - 1 x 2 = 1 Thus lim k →∞ a k = lim x →∞ f ( x ) = e lim x →∞ x ln(1 - 1 x 2 ) = e . Theorem 2. If a = 0 , and r C then lim k →∞ ar k = 0 if | r | < 1; a if r = 1 diverges otherwise 1

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Proof. Set f ( x ) = a | r | x = ae x ln | r | . If | r | < 1 then ln | r | < 0 and lim x →∞ ae x ln | r | = 0. Thus lim k →∞ ar k = 0. If | r | > 1 then ln | r | > 0 and lim x →∞ ae x ln | r | = ±∞ so ar k diverges. If r = 1 then ar k = a converges to a .
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