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Math Help - Limit with Factorials

  1. #1
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    Limit with Factorials

    Hi. I need to find the limit as m goes to infinity of this function:

    [\frac{m!m!}{(m+k)!(m-k)!}]^m. It might be sort of hard to read but it's that whole fraction raised to the m power.

    My teacher first suggested trying to compute the limit as m goes to infinity of [\frac{m}{m+k}]^m. I still wasn't exactly sure even how to do that, but tried using L'Hospital and changed it to the limit as m goes to infinity of \frac{(ln(m) + 1)m^m}{(ln(m+k)+\frac{m}{m+k})(m+k)^m}, which honestly made me even more confused.

    THEN, she said to use the fact that a = e^{ln(a)} and set a=\frac{m}{m+k}. All in all, I just feel like I'm getting even more jumbled up!

    The book says that the limit for the original equation (up at the top) should be e^{-k^2}. Can anyone help me out?
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  2. #2
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    Ok, after thinking more about it, I think I've made a little more sense of what my teacher was trying to say. Instead of computing the limit as m goes to infinity of [\frac{m}{m+k}]^m, she is saying to computing the limit as m goes to infinity of e^{m ln(\frac{m}{m+k})}. But at this point, I'm still confused on how to do that.
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  3. #3
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    Quote Originally Posted by azdang View Post
    Ok, after thinking more about it, I think I've made a little more sense of what my teacher was trying to say. Instead of computing the limit as m goes to infinity of [\frac{m}{m+k}]^m, she is saying to computing the limit as m goes to infinity of e^{m ln(\frac{m}{m+k})}. But at this point, I'm still confused on how to do that.
    I am not sure that this helps. But it is correct.
    \left( {\frac{m}<br />
{{m + k}}} \right)^m  = \left( {1 - \frac{k}<br />
{{m + k}}} \right)^m  \to e^{ - k}
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  4. #4
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    Quote Originally Posted by azdang View Post
    Hi. I need to find the limit as m goes to infinity of this function:

    [\frac{m!m!}{(m+k)!(m-k)!}]^m. It might be sort of hard to read but it's that whole fraction raised to the m power.

    My teacher first suggested trying to compute the limit as m goes to infinity of [\frac{m}{m+k}]^m. I still wasn't exactly sure even how to do that, but tried using L'Hospital and changed it to the limit as m goes to infinity of \frac{(ln(m) + 1)m^m}{(ln(m+k)+\frac{m}{m+k})(m+k)^m}, which honestly made me even more confused.

    THEN, she said to use the fact that a = e^{ln(a)} and set a=\frac{m}{m+k}. All in all, I just feel like I'm getting even more jumbled up!

    The book says that the limit for the original equation (up at the top) should be e^{-k^2}. Can anyone help me out?
    if k = 0, the limit is obviously 1. so i'll assume that k > 0. we have \frac{m!}{(m+k)!}=\frac{1}{(m+k)(m+k-1) \cdots (m+1)} and \frac{m!}{(m-k)!}=m(m-1) \cdots (m-k+1). thus:

    I_m=\left(\frac{m!m!}{(m+k)!(m-k)!} \right)^m=\left(\frac{m}{m+k} \right)^m \left(\frac{m-1}{m+k-1} \right)^m \cdots \left(\frac{m-k+1}{m+1} \right)^m, which can also be written as: I_m=\prod_{j=0}^{k-1} \left(\frac{m-j}{m+k-j} \right)^m. \ \ \ \ \ \ (1)

    now \left(\frac{m-j}{m+k-j} \right)^m=e^{m[\ln(m-j) - \ln(m+k-j)]} and using L'Hopital rule we have: \lim_{m\to\infty} m[\ln(m-j) - \ln(m+k-j)]=\lim_{m\to\infty}\frac{\ln(m-j)- \ln(m+k-j)}{\frac{1}{m}}=-k.

    therefore \lim_{m\to\infty}\left(\frac{m-j}{m+k-j} \right)^m = e^{-k}, for all 0 \leq j \leq k-1, and hence by (1): \ \ \lim_{m\to\infty}I_m=e^{-k^2}.
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  5. #5
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    This is EXTREMELY helpful! Thank you so much, NonCommAlg! My teacher had more or less given me a bunch of hints but I just couldn't put them alltogether as you have

    I am still a bit confused on where the -k^2 comes from as the exponent rather than just -k. I don't really see it. What am I missing?
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  6. #6
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    Quote Originally Posted by azdang View Post
    This is EXTREMELY helpful! Thank you so much, NonCommAlg! My teacher had more or less given me a bunch of hints but I just couldn't put them alltogether as you have

    I am still a bit confused on where the -k^2 comes from as the exponent rather than just -k. I don't really see it. What am I missing?
    I_m is the product of k factors (see (1) again) and the limit of each factor, as i showed, is e^{-k}. so the limit of I_m is e^{-k} \cdot e^{-k} \cdots \cdot e^{-k}, k times, which is equal to e^{-k^2}.
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  7. #7
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    Oh duh! Thank you for the clarification. I really appreciate it.
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  8. #8
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    Or without L'Hospital

    \mathop {\lim }\limits_{m \to \infty } {\left( {\frac{m - j}{m + k - j}} \right)^m} = \mathop {\lim }\limits_{m \to \infty } {\left(1 + \frac{-k}{m + k - j}\right)^m} = \left\{ \begin{gathered}\frac{- k}{m + k - j} = \frac{1}{n}, \hfill \\ m =  - kn - \left( {k - j} \right) \hfill \\m \to \infty , \, n \to -\infty  \hfill \\ \end{gathered}  \right\} =

    = \mathop {\lim }\limits_{n \to -\infty} {\left(1 + \frac{1}{n}\right)^{ - kn - \left( {k - j} \right)}} = \mathop {\lim }\limits_{n \to -\infty } {\left[ {{{\left(1 + \frac{1}{n}\right)}^n}} \right]^{-k}}{\left(1 + \frac{1}{n}\right)^{ - (k - j)}}=

    =e^{-k} \cdot {1^{-(k - j)}} = e^{-k}.
    Last edited by DeMath; October 4th 2009 at 06:13 AM. Reason: typo
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