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Math Help - limit to log

  1. #1
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    limit to log

    Could someone show me how to solve the following limits algebraically:

    For x > 1, lim as n approaches infinity of

    a) [(1 + x^(1/n))/2]^(2n) = x

    b) [(1 + x^(1/n))/2]^(n) = sqrt(x)

    They seem very similar to the sequence (1 + (a/n))^n whose limit is e^a, but I don't see how to manipulate it accordingly.

    If you could show me how to solve these, I would really appreciate it.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by BrainMan View Post
    Could someone show me how to solve the following limits algebraically:

    For x > 1, lim as n approaches infinity of

    a) [(1 + x^(1/n))/2]^(2n) = x

    b) [(1 + x^(1/n))/2]^(n) = sqrt(x)

    They seem very similar to the sequence (1 + (a/n))^n whose limit is e^a, but I don't see how to manipulate it accordingly.

    If you could show me how to solve these, I would really appreciate it.

    Thanks in advance.
    What are the "=" signs doing here? Is x changing with n?

    RonL
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  3. #3
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    Quote Originally Posted by BrainMan View Post
    Could someone show me how to solve the following limits algebraically:

    For x > 1, lim as n approaches infinity of

    a) [(1 + x^(1/n))/2]^(2n) = x

    b) [(1 + x^(1/n))/2]^(n) = sqrt(x)

    They seem very similar to the sequence (1 + (a/n))^n whose limit is e^a, but I don't see how to manipulate it accordingly.

    If you could show me how to solve these, I would really appreciate it.

    Thanks in advance.
    You can write,
    (1+x^{1/n})/2 = 1+a_n/n where a_n = n(x^{1/n}-1)/2. So now you need to compute (1+a_n/n)^{1/n} \to e^a where a=\lim a_n. Thus, it remains to find n(x^{1/n}-1)\to \log x
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    What are the "=" signs doing here? Is x changing with n?

    RonL
    That's what the limits equal. Sorry about the confusion.

    I need the limit of [(1 + x^(1/n))/2]^(2n), which equals x, and the limit of [(1 + x^(1/n))/2]^(n), which equals the square root of x.

    I just don't know how to show them with algebra.

    PerfectHacker, thanks for the response, but I don't really see what you're doing. Could you elaborate? Basically, how does it simplify to x?
    Last edited by BrainMan; October 24th 2007 at 11:47 AM.
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  5. #5
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    Ignore the last post. Looking at it more closely, I see what you did. However, how do you know the last limit goes to ln(x)? Could you just explain that?
    Last edited by BrainMan; October 24th 2007 at 11:58 AM.
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  6. #6
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    Quote Originally Posted by BrainMan View Post
    Ignore the last post. Looking at it more closely, I see what you did. However, how do you know the last limit goes to ln(x)? Could you just explain that?
    Yes. This limit is all that remains.

    x^{1/n} - 1 = [1+(x-1)]^{1/n} - 1 = \frac{1}{n}(x-1) + \frac{1}{n} \left( \frac{1}{n} - 1 \right)\frac{(x-1)^2}{2!}+ \frac{1}{n} \left(\frac{1}{n} - 1 \right) \left( \frac{1}{n} - 2\right) \frac{(x-1)^3}{3!}+...

    So n(x^{1/n}-1) is,
    (x-1) + \left( \frac{1}{n} - 1 \right)\frac{(x-1)^2}{2!}+  \left(\frac{1}{n} - 1 \right) \left( \frac{1}{n} - 2\right) \frac{(x-1)^3}{3!}+...
    As we take the limit we end up with,
    \frac{(x-1)}{1} - \frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-... = \log x

    This is Mine 74th Post!!!
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