But parenleftBig 1 x n parenrightBig n e x cadena jc59484 HW10 lawn 55930 3 as

# But parenleftbig 1 x n parenrightbig n e x cadena

This preview shows page 2 - 6 out of 11 pages.

But parenleftBig 1 + x n parenrightBig n −→ e x
cadena (jc59484) – HW10 – lawn – (55930) 3 as n . Consequently, { a n } converges and has limit = ( e 1 ) 6 = e 6 . 006 10.0points Determine whether the sequence { a n } con- verges or diverges when a n = ( 1) n parenleftbigg 6 n + 3 7 n + 7 parenrightbigg , and if it does, find its limit. 1. sequence diverges correct 2. limit = 3 7 3. limit = ± 6 7 4. limit = 6 7 5. limit = 0 Explanation: After division, 6 n + 3 7 n + 7 = 6 + 3 n 7 + 7 n . Now 3 n , 7 n 0 as n → ∞ , so lim n → ∞ 6 n + 3 7 n + 7 = 6 7 negationslash = 0 . Thus as n → ∞ , the values of a n oscillate be- tween values ever closer to ± 6 7 . Consequently, the sequence diverges . 007 10.0points Determine if the sequence { a n } converges, when a n = 4 cos parenleftbigg 2 n π + 1 6 n + 7 parenrightbigg , and if it does, find its limit. 1. limit = 2 3 2. limit = 4 3 π 3. limit = cos 1 7 4. limit = 4 cos 1 7 5. limit = 2 correct 6. sequence does not converge Explanation: After division, 2 n π + 1 6 n + 7 = 2 π + 1 n 6 + 7 n . But 1 n , 7 n 0 as n → ∞ , so lim n → ∞ 2 n π + 1 6 n + 7 = π 3 . Consequently, since cos x is continuous as a function of x , the sequence { a n } converges and has limit = 4 cos π 3 = 2 . 008 10.0points Determine if the sequence { a n } converges when a n = (2 n + 1)! 7 n 2 (2 n 1)! , and if it converges, find the limit. 1. does not converge
cadena (jc59484) – HW10 – lawn – (55930) 4 2. converges with limit = 7 2 3. converges with limit = 4 7 correct 4. converges with limit = 2 7 5. converges with limit = 7 4 Explanation: By definition, m ! is the product m ! = 1 · 2 · 3 · . . . · m of the first m positive integers. When m = 2 n 1, therefore, (2 n 1)! = 1 · 2 · 3 · . . . · (2 n 1) , while (2 n + 1)! = 1 · 2 · 3 · . . . · (2 n 1)2 n (2 n + 1) . when m = 2 n + 1. But then, (2 n + 1)! 7 n 2 (2 n 1)! = 2 n (2 n + 1) 7 n 2 −→ 4 7 as n → ∞ . Consequently, the given sequence converges with limit = 4 7 . 009 10.0points Which of the following sequences converge? A . braceleftbigg 5 e n 2 + e n bracerightbigg B . braceleftbigg e n + 5 6 n + 4 bracerightbigg 1. both A and B 2. B only 3. neither of them 4. A only correct Explanation: A . After simplification, 5 e n 2 + e n = 5 2 e n + 1 . Thus 5 e n 2 + e n −→ 5 as n → ∞ since e x 0 as x → ∞ . B . To test for convergence of the sequence braceleftbigg e n + 5 6 n + 4 bracerightbigg we use L’Hospital’s Rule with f ( x ) = e x + 5 , g ( x ) = 6 x + 4 since f ( x ) −→ ∞ , g ( x ) −→ ∞ as x → ∞ . Thus lim n → ∞ e n 6 n + 4 = lim x → ∞ f ( x ) g ( x ) . But f ( x ) g ( x ) = e x 6 −→ ∞ as n → ∞ . Consequently, only A converges . 010 10.0points Determine whether the sequence { a n } con- verges or diverges when a n = ln(5 n 6 ) ln(3 n 5 ) , and if it converges, find the limit. 1. diverges 2. converges with limit = 0
cadena (jc59484) – HW10 – lawn – (55930) 5 3. converges with limit = 6 5 correct 4. converges with limit = ln(5) ln(3) 5. converges with limit = 5 3 Explanation: By properties of logs, ln(5 n 6 ) = ln(5) + 6 ln( n ) , ln(3 n 5 ) = ln(3) + 5 ln( n ) .

#### You've reached the end of your free preview.

Want to read all 11 pages?

• Fall '11
• Gramlich
• Accounting, Limit, lim