Calculus 2

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Find the limit as x approaches infinity of (1-(1/(2x)))^(2x).

  • Calculus 2 -

    f(x)= [1-1/(2x)]^(2 x)

    Log[f(x)] = 2x Log[1-/(2x)] =

    2 x [1/(2x) + O(1/x^2)] =

    1 + O(1/x)

    The limit for x to infinity of
    Log[f(x)] is thus 1. The limit for x to infinity of f(x) is thus e, because the logarithmic function is continuous there.

  • Calculus 2 -

    Hmmm. Let u = 1/2x and you have

    (1+u)^(-1/u) = 1 / (1+u)^(1/u)
    as x->infinity, u->0, and we have

    1/e

  • Calculus 2 -

    I see, I forgot a minus sign.



    f(x)= [1-1/(2x)]^(2 x)

    Log[f(x)] = 2x Log[1-/(2x)] =

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

    -1 + O(1/x)

    So, you get exp(-1), using the fact that log is continuous there.

  • Calculus 2 -

    thank you! we are supossed to use Lopital's rule though. Is there a different way to do it using Lopital's?

  • Calculus 2 -

    If you have to use L'Hôpital, then that's close to how I did it above. After taking logarithms, you have:

    Log[f(x)] = 2x Log[1-/(2x)]

    You can write this as:

    Log[1-1/(2x)] / (1/(2x))

    This is then a 0/0 case. If you put
    1/(2x) = u then this becomes:

    Log[1-u]/u

    and you need to compute the limit of u to zero.

  • Calculus 2 -

    thank you so much!

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