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Math Help - L'hopitals rule

  1. #1
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    L'hopitals rule

    Lim (x-->0) [2-e^3x]^1/x = I.F 1^Inf.

    Lim (x-->0) 1/x Ln(2-e^3x)= I.F. Inf. * 0

    Lim(x-->0) x^-1 Ln (2-e^3x) = Ln (2-e^3x)/(1/x^-1)
    I don't think I did that last step right. a little insight would be great. Thanks
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  2. #2
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    i think i got it .

    Lim (Ln(2-e^3x)/x = (3e^3x)/2-e^3x = 3/1 = 3 ... is that right?
    x-->0
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