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Winter 2007 - Fitzsimmons' Class - Exam 1

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

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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 ǫ > 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 < ǫ . That is (2 . 2) n n * = 1 n < ǫ. Consequently, by (2.1) and (2.2), n n * = vextendsingle vextendsingle vextendsingle vextendsingle n n + n - 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ, as required.

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