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asst6soln - MATH 138 Calculus 2 Solutions to Assignment 6...

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MATH 138 Calculus 2, Solutions to Assignment 6 Problem 1: We can sense that ( 3 + 1 n ) 2 9 as n → ∞ . Given any > 0, show that if n > 7 , then ( 3 + 1 n ) 2 - 9 < . This proves from scratch that ( 3 + 1 n ) 2 9. Hint. You know that 1 n 2 1 n always. Solution: For n 1 we have n 2 n so 1 n 2 1 n , and we have ( 3 + 1 n ) 2 > (3 + 0) 2 = 9 so ( 3 + 1 n ) 2 - 9 = ( 3 + 1 n ) 2 - 9 = ( 9 + 6 n + 1 n 2 ) - 9 = 6 n + 1 n 2 6 n + 1 n = 7 n . Thus given > 0, if n > 7 then 7 n < so we have ( 3 + 1 n ) 2 - 9 7 n < . Problem 2: Find the following sequence limits by using limit properties. It is not necessary to prove your answers by means of the -definition. (a) lim n →∞ n q 1 + 1 n - 1 Solution 1: We have lim n →∞ n q 1 + 1 n - 1 = lim n →∞ n q 1 + 1 n - 1 q 1 + 1 n + 1 q 1 + 1 n + 1 = lim n →∞ n ( 1 + 1 n - 1 ) q 1 + 1 n + 1 = lim n →∞ 1 q 1 + 1 n + 1 = 1 1 + 0 + 1 = 1 2 . Solution 2: By replacing 1 n by the continuous variable x and then using l’Hˆ opital’s Rule, we have lim n →∞ n q 1 + 1 n - 1 = lim n →∞ q 1 + 1 n - 1 1 n = lim x 0 + 1 + x - 1 x = lim x 0 + 1 2 1+ x 1 = 1 2 . (We remark that when we replace 1 n by x , we are implicitly using Theorem 12 from the notes on limits of sequences which are posted on UW-ACE). (b) lim n →∞ n 3 q 1 + 1 n - 1 . This one might require more imagination. Solution 1: We have lim n →∞ n 3 q 1 + 1 n - 1 = lim n →∞ n ( 1 + 1 n ) 1 / 3 - 1 ( 1 + 1 n ) 2 / 3 + ( 1 + 1 n ) 1 / 3 + 1 ( 1 + 1 n ) 2 / 3 + ( 1 + 1 n ) 1 / 3 + 1 = lim n →∞ n ( 1 + 1 n - 1 ) ( 1 + 1 n ) 2 / 3 + ( 1 + 1 n ) 1 / 3 + 1 = lim n →∞ 1 ( 1 + 1 n ) 2 / 3 + ( 1 + 1 n ) 1 / 3 + 1 = 1 (1 + 0) 2 / 3 + (1 + 0) 1 / 3 + 1 = 1 3 . Solution 2: By replacing 1 n by the continuous variable x and then using l’Hˆ opital’s Rule, we have lim n →∞ n 3 q 1 + 1 n - 1 = lim n →∞ ( 1 + 1 n ) 1 / 3 - 1 1 n = lim x 0 + (1 + x ) 1 / 3 - 1 x = lim x 0 + 1 3 (1 + x ) - 2 / 3 1 = 1 3 . 1
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(c) lim n →∞ (ln(3 + e n )) /n Hint: First explain and then use the identity ln(3 + e n ) = n + ln (1 + 3 e - n ) . Solution 1: Note that ln(3 + e n ) = ln ( e n ( 3 e n + 1 )) = ln ( e n ) + ln ( 3 e n + 1 ) = n + ln (1 + 3 e - n ), so lim n →∞ ln(3 + e n ) n = lim n →∞ n + ln (1 + 3 e - n ) n = lim n →∞ 1 + 1 n ln (1 + 3 e - n ) = 1 + 0 · ln(1 + 3 · 0) = 1 . Solution 2: By replacing n by the continuous variable x then using l’Hˆ opital’s Rule, we have lim n →∞ ln(3 + e n ) n = lim x →∞ ln(3 + e x ) x = lim x →∞ e x 3+ e x 1 = lim x →∞ 1 3 e - x + 1 = 1 3 · 0 + 1 = 1 .
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