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Math Help - Limit Problem

  1. #1
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    Limit Problem

    Hi, I'm trying to work out (limit as n=>infinity) (n * (sqrt(1 + 1/n) - 1)).

    I know that by letting n = 1/t and then using l'hopitals this equals 1/2.
    However, I need to use only basic limit rules (add/sub/mult/div and composition).

    I feel like I am really close to the answer because I have been able to rewrite this as (n * (1 + 1/n)^(1/2)) - n.

    I noted that:
    ln(n * (1 + 1/n)^(1/2))) - ln(n)
    = ln((1 + 1/n)^(1/2))
    = 1/2(ln(1+1/n))
    and since lim(1+1/n) = e as n=>inf
    = 1/2(ln(e))
    = 1/2

    I cannot figure out any way to actually bring the ln into the equation like that though...any thoughts? Suggestions would be much appreciated.
    Last edited by crymorenoobs; February 25th 2010 at 08:14 PM.
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  2. #2
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    Hello crymorenoobs

    Welcome to Math Help Forum!
    Quote Originally Posted by crymorenoobs View Post
    Hi, I'm trying to work out (limit as n=>infinity) (n * (sqrt(1 + 1/n) - 1)).

    I know that by letting n = 1/t and then using l'hopitals this equals 1/2.
    However, I need to use only basic limit rules (add/sub/mult/div and composition).

    I feel like I am really close to the answer because I have been able to rewrite this as (n * (1 + 1/n)^(1/2)) - n.

    I noted that:
    ln(n * (1 + 1/n)^(1/2))) - ln(n)
    = ln((1 + 1/n)^(1/2))
    = 1/2(ln(1+1/n))
    and since lim(1+1/n) = e as n=>inf
    = 1/2(ln(e))
    = 1/2

    I cannot figure out any way to actually bring the ln into the equation like that though...any thoughts? Suggestions would be much appreciated.
    What about the Binomial Expansion?
    \left(1+\frac1n\right)^{\frac12}
    =1+\left(\frac12\right)\left(\frac1n\right)+\frac{  1}{2!}\left(\frac12\right)\left(-\frac12\right)\left(\frac1n\right)^2+\frac{1}{3!}\  left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)\left(\frac1n\right)^3+...
    Subtract 1, multiply by n, and you're there.

    Grandad
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  3. #3
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    Oohhhhhh, that would do it. Thanks so much.
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  4. #4
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    Unfortunately, ln(a- b) is NOT equal to ln(a)- ln(b) so your first sentence using ln is incorrect.
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  5. #5
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    Yeah, I was wondering if there was maybe a trick I didn't know about or something because that was the only way I could see to incorporate 1/2..by bringing down the exponent. Guess not though :P
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  6. #6
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     <br />
\lim_{n \to \infty} n\left[\sqrt{1+\frac{1}{n}} -1\right] \cdot \frac{\sqrt{1+\frac{1}{n}} +1}{\sqrt{1+\frac{1}{n}}+1}<br />

     <br />
\lim_{n \to \infty} n \cdot \frac{\left(1+\frac{1}{n}\right) - 1}{\sqrt{1+\frac{1}{n}}+1}<br />

     <br />
\lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{2}<br />
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