calculus

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take the limit as x goes to infinity: (14x/(14x+3))^(4x)

  • calculus -

    x = 1
    f = (14/17)^4 = .46

    x = 10
    f = (140/143)^40 = .428

    x = 100
    f = (1400/1403)^400 = .4248

    x = 1000
    f = (14,000/14,003)^4000 = .4244

    hmmm

  • calculus -

    x = 10,000
    f = (140,000/140,003)^40,000 = .4244
    double hmmm

  • calculus -

    x = 100,000
    f = (1,400,000/1,400,003)^400,000 = .4244

  • calculus -

    Log[f(x)] =

    4 x [log(14 x) - log(14 x + 3)] =

    -4 x log[1 + 3/(14 x)] =

    -4 x [3/(14 x) + O(1/x^2)] =

    -6/7 + O(1/x)

    The limit for x to infinity of
    log[f(x)] is thus -6/7, the limit of
    f(x) is thus exp(-6/7)

  • calculus -

    What happened between
    -4 x log[1 + 3/(14 x)] =
    and
    -4 x [3/(14 x) + O(1/x^2)] =
    How did you get rid of the log? Thx in advance

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