Winter 2007 - Fitzsimmons' Class - Exam 1

Winter 2007 - Fitzsimmons' Class - Exam 1 - Math 142A (P....

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Unformatted text preview: Math 142A (P. Fitzsimmons) First Midterm Exam Solutions 2. Use the definition of limit (of a sequence) to show that lim n n n + n = 1 . Solution. Let us begin by estimating the distance from the sequence to the limit: (2 . 1) vextendsingle vextendsingle vextendsingle vextendsingle n n + n- 1 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle n- n- n n + n vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle- n n + n vextendsingle vextendsingle vextendsingle vextendsingle = n n + n n n = 1 n . Given > 0 (and with the above calculation in mind) we choose the cutoff n * to be any integer larger that 1 / 2 . If n n * , then n > 1 /e 2 , so n > 1 / , so 1 / n < . That is (2 . 2) n n * = 1 n < ....
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This note was uploaded on 04/23/2008 for the course MATH 142A taught by Professor Fitzsimmons during the Winter '01 term at UCSD.

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Winter 2007 - Fitzsimmons' Class - Exam 1 - Math 142A (P....

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