# Math Help - Limit with Factorials

1. ## 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?

2. 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.

3. Originally Posted by azdang
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.
$\left( {\frac{m}
{{m + k}}} \right)^m = \left( {1 - \frac{k}
{{m + k}}} \right)^m \to e^{ - k}$

4. Originally Posted by azdang
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 $(1): \ \ \lim_{m\to\infty}I_m=e^{-k^2}.$

5. 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?

6. Originally Posted by azdang
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 $e^{-k^2}.$

7. Oh duh! Thank you for the clarification. I really appreciate it.

8. Or without L'Hospital

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

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

$=e^{-k} \cdot {1^{-(k - j)}} = e^{-k}.$