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I need to find the limit of this function..I thought about L'hôpital's rule, but can't seem to derive them both..

$$\lim_{n\rightarrow\infty} \frac{(2n)!}{(n!)^2}$$

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Are you allowed to use Stirling approximation of $n!$ ? – Claude Leibovici Jun 21 '14 at 11:45
Nothing said about that..but if I dosent want to use it, there is a way? – Pili66 Jun 21 '14 at 11:47
Yes. Cancel an $n!$ from top and bottom, then pair up the remaining terms. You will see that each resulting fraction is at least one, and the last one is $\frac{n+1}{1}$, so that the product is at least $n+1$. – rogerl Jun 21 '14 at 11:49
You are taking the limit of an increasing sequence of integer numbers. Can you guess what the limit is? – Jack D'Aurizio Jun 21 '14 at 12:26

Hint: Note that $$\frac{(2n)!}{n!n!}=\frac{(2n)(2n-1)(2n-2)\cdots (n+1)}{(n)(n-1)(n-2)\cdots (1)}.$$ Each of the $n$ fractions $\frac{2n}{n}$, $\frac{2n-1}{n-1}$, $\frac{2n-2}{n-2}$, and so on down to $\frac{n+1}{1}$ is $\ge 2$. So the thing blows up quite fast.

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So finaly the limit is some const or infinty? – Pili66 Jun 21 '14 at 11:53
Language used in courses is not uniform. Some would say the limit is infinite, or $\infty$. Some would say the limit does not exist. If $\infty$ is allowed in your course as an answer, then that is the limit. – André Nicolas Jun 21 '14 at 11:56

Let us assume that you use Stirling approximation which is $$p! \simeq \sqrt {2 \pi p} \Big(\frac{p}{e}\Big)^p$$ So, applying it to your terms, you get $$I_n=\frac{(2n)!}{(n!)^2}=\frac{2^{2 n}}{\sqrt{\pi n}} $$ which, as nicely demonstrated by André Nicolas in his answer, increases very fast (if $n=10$, the exact value of the expression is $184756$ while the approximation I used gives $187079$).

This solution is probably off-topic to you but just remember this approximation of $p!$; for sure you will be using a lot in your studies.

Added later

There is even a simpler way using logarithms and Stirling first approximation $$\log(p!) \simeq p \log(p)-p$$ So, in your case $$\log(I_n) \simeq \Big(2n \log(2n)-2n\Big)-2 \Big(n \log(n)-n\Big)=2n\log(2)$$ So, $I_n \simeq2^{2n}$.

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