midterm_21c_solutions

midterm_21c_solutions - Calculus Math 21C Fall 2010 Midterm...

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Unformatted text preview: Calculus Math 21C, Fall 2010 Midterm I: Solutions 1. [20%] Do the following sequences { a n } converge or diverge as n → ∞ ? If a sequence converges, find its limit. Justify your answers. (a) a n = 2 + (- 1) n ; (b) a n = n e n ; (c) a n = 1 + 2 n n . Solution. • (a) The sequence diverges since it oscillates between 1 and 3. For example, if = 1, there is no number L such that | a n- L | < for all sufficiently large n , since then we would have both | 1- L | < 1 (or < L < 2) and | 3- L | < 1 or (2 < L < 4), which is impossible. So there is no L that satisfies the definition of the limit. • (b) Both the numerator n and the denominator e n are differentiable functions that diverge to ∞ as n → ∞ . Differentiating the numerator and denominator with respect to n , we get by l’Hˆ ospital’s rule that lim n →∞ n e n = lim n →∞ 1 e n = 0 . Thus, the sequence converges to 0. • (c) The sequence converges to e 2 . To show this, let b n = ln a n = n ln 1 + 2 n . Then, writing n = 1 /x , we have lim n →∞ b n = lim x → ln(1 + 2 x ) x . The fraction is an indeterminate limit (0 / 0) of differentiable functions, so by l’Hˆ ospital’s rule lim x → ln(1 + 2 x ) x = lim x → 2 / (1 + 2 x ) 1 = lim x → 2 1 + 2 x = 2 . 1 Thus, b n → 2 as n → ∞ . Since a n = e b n and e x is a continuous function, it follows that lim n →∞ a n = e lim n →∞ b n = e 2 . • More generally, for any-∞ < x < ∞ , we have the limit e x = lim n →∞ 1 + x n n 2 2. [20%] Determine whether the following series converge or diverge. You can use any appropriate test provided that you explain your answer....
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This note was uploaded on 11/17/2010 for the course MAT 21C NA taught by Professor Na during the Fall '01 term at UC Davis.

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midterm_21c_solutions - Calculus Math 21C Fall 2010 Midterm...

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