exam2sol - MAT 127: Calculus C, Fall 2010 Solutions to...

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Unformatted text preview: MAT 127: Calculus C, Fall 2010 Solutions to Midterm II Problem 1 (20pts) Answer Only: no explanation is required. Write your answer to each question in the correspond- ing box in the simplest possible form. No credit will be awarded if the answer in the box is wrong; partial credit may be awarded if the answer in the box is correct, but not in the simplest possible form. In (a)-(c), assume that the limits exist. (a; 5pts) Find the limit of the sequence a n = ln(64 n 2 + 1) ln( n 2 + n ) 4 3 2 ln 2 This is similar to 8.1 33 (on HW5). Since a n = 1 4 ln parenleftbigg 64 n 2 + 1 n 2 + n parenrightbigg = 1 4 ln parenleftbigg 64 n 2 /n 2 + 1 /n 2 n 2 /n 2 + n/n 2 parenrightbigg = 1 4 ln parenleftbigg 64 + 1 /n 2 1 + n/n 2 parenrightbigg , plugging in n = we obtain a n 1 4 ln parenleftbigg 64 + 1 / 1 + 1 / parenrightbigg = 1 4 ln parenleftbigg 64 + 0 1 + 0 parenrightbigg = 1 4 ln 2 6 = 6 4 ln 2 = 3 2 ln 2 . Grading: wrong answer 0pts; 1 4 ln 64 2pts; 1 2 ln 8 4pts; as above 5pts (b; 5pts) Find the limit of the sequence a n = n ( 1 e 1 /n ) 1 This is similar to 8.1 23,27 (the latter on HW5; the former was a suggested practice problem). lim n a n = lim n 1 e 1 /n 1 /n = lim x 1 e x x = lim x e x 1 = e = 1 . The third equality uses lHospital, which is applicable here because (1 e x ) , x 0 as x 0. Grading: wrong answer 0pts; as above 5pts (c; 5pts) Find the limit of the sequence 5 15 , radicalBig 15 + 2 15 , radicalbigg 15 + 2 radicalBig 15 + 2 15 , radicalBigg 15 + 2 radicalbigg 15 + 2 radicalBig 15 + 2 15 , . . . This is similar to 8.1 54 (on HW6). Since a n +1 = 15 + 2 a n , if this sequence converges to a , then a = 15 + 2 a = a 2 = 15 + 2 a = a 2 2 a 15 = 0 = ( a 5)( a + 3) = 0 . Since a 0 by the first statement, we conclude that a = 5. Grading: wrong answer 0pts; as above 5pts (d; 5pts) Write the number 1 . 1 45 = 1 . 1454545 . . . as a simple fraction 63 55 This is similar to 8.2 36,38 (on HW6): 1 . 1 45 = 1 . 1 + . 045 + . 045 1 100 + . 045 1 100 2 + . . . = 11 10 + 45 / 1000 1 1 100 = 11 10 + 45 / 10 99 = 11 10 + 5 110 = 121 + 5 110 = 63 55 Grading: wrong answer 0pts; as above 5pts; 1 8 55 or not simplified 4pts; both issues 3pts Problem 2 (15pts) Determine whether each of the following sequences converges or diverges. In each case, circle your answer to the right of the question and justify it in the space provided below the question. You do not have to determine the limit if the sequence converges. (a; 5pts) a n = ( 1) n n ! n n (reminder: n ! = 1 2 . . . n ) converge diverge Since n ! /n n 0 (by Example 9 on p558), a n 0 (by Theorem 4 on p557). Alternatively, the Ratio Test for Sequences can be used: | a n +1 | | a n | = ( n +1)! / ( n +1) n +1 n ! /n n = ( n +1) n n ( n +1) n +1 = parenleftbigg n n + 1 parenrightbigg n = 1 (1 + 1 /n ) n = 1 e ....
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exam2sol - MAT 127: Calculus C, Fall 2010 Solutions to...

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