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December 18, 2014

December 18, 2014

Posted by **Roy** on Tuesday, April 6, 2010 at 8:22am.

I'm thinking that you have to use the squeeze theorem, but I'm not quite sure how. help please?

- calculus -
**Reiny**, Tuesday, April 6, 2010 at 9:24amhow about something like this ....

n!/(2n)!

= n(n-1)(n-2)(n-3) ... (2)(1)/ [(2n)(2n-1)(2n-2)(2n-3) ..(2)(1) ]

= n(n-1)(n-2)(n-3) ... (2)(1)/ 2n(2n-1)(2)(n-1)(2n-3)(2)(n-2)(2n-5)(2)(n-3) ... (2)(1) ]

notice that all of the factors of the numerator are found in the denominator, so ...

= 1/ [ 2^n(2n-1)(2n-3)(2n-5) ... (2)(1) ]

clearly as n ---> ∞, the denominator --- ∞

1/(very large) ---> 0

and the limit n!/(2n)! = 0

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