maths-calculus

find the limit as x approaches infinity, -(x+1) (e^(1/(x+1))-1)

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asked by bel
  1. If you let u = -(x+1), you have

    u(e^(-1/u) - 1)
    = (e^(-1/u) - 1) / (1/u)

    Now using l'Hospital's Rule, that is the same as

    (e^(-1/u) - 1)(1/u^2) / (1/u^2)
    = e^(-1/u) - 1
    -> 0-1 = -1

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    posted by Steve

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