Unformatted text preview: n â†’âˆž 2 n + 1 n 2 + n 2 n1 n 2n = lim n â†’âˆž 2 n + 1 2 n1 Â· n 2n n 2 + n = 1 since the leading term on top and bottom is now 2 n 3 . (7) This limit will be 0. If we apply Lâ€™HË†opitalâ€™s Rule repeatedly we end up with 18 24 n which goes to 0 as n goes to inï¬nity. (8) This limit exists and equals1. (Recall arctan n goes to Ï€/ 2 as n goes to inï¬nity.) (9) Since (1 + 1 /n ) n goes to e as n goes to inï¬nity, this will go to e 2 . (10) (a) The numerator goes to e and the denominator goes to âˆž . So the quotient will go to 0 as n goes to âˆž . (b) An âˆž âˆž form so use Lâ€™HË† opitalâ€™s Rule to show that the limit is 1 / 2. 1...
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 Fall '07
 Nelson
 Calculus, Geometric Series, lâ€™HË†pitalâ€™s rule

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