Thursday

October 23, 2014

October 23, 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

**Answer this Question**

**Related Questions**

math - Prove that the limit as n approaches infinity of ((n!)^2)/(2n)! is 0. I'm...

calculus - Suppose that f(x) is bounded: that is, there exists a constant M ...

Calculus - I'm supposed to find the limit as x approaches infinity of (2-x-sinx...

calculus - It is known that x 2 + 4x ≤ f(x) ≤ -x 2 -4x the interval...

calculus - Suppose that f(x) is bounded: that is, there exists a constant M such...

Math-Calculus - Hi, I am trying to figure out what the limit as h approaches 0 ...

Calculus - Following 2 questions are from a book at a point where L’Hopital’s ...

Calculus - Following 2 questions are from a book at a point where L’Hopital’s ...

Help with one limit - I'm asked to find the limit as n approaches infinity of...

calculus - Lim sin2h sin3h / h^2 h-->0 how would you do this ?? i got 6 as ...