Value of limit as factorial approaches infinity?

Question

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Here's the problem

As n approaches infinity what is

(n+3)!(2(n-1))!
-------------------
n!(2n+1)!

I got 1 as an answer but it can't be?

It's either
a) 1/8
b) -5/8
c) -1/8
d) 1/4
e) none of the above

I've already eliminated the n! and was left with

(n+3)(2n-2)
---------------
(2n+1)

I suppose here's where I start to err?

Answer 1

(n+3)! = (n+3) (n+2) (n+1) n!
(2n+1)! = (2n+1) 2n (2n-1) (2n-2)!

So you can simplify by n! and (2n-2)!

So (n+3) (n+2) (n+1)/ (2n+1) 2n (2n-1)

So at infinite n is more important than number so
it is n n n/ 2n2n2n = 1/8

1/8 is the answer answer a

Answer 2

[(n+3)!] * [(2(n-1))!] / [n!] * [(2n+1)!]

Try reducing the factorials first:
For example, if you had

[n!] / [(n - 1)!] = n
because all the factors prior to "n" in the "[n!]" expression are just (n - 1)*(n - 2)*(n - 3)*... etc. which is just [(n - 1)!]

[(2n + 1)!] / [(2n - 2)!] = (2n + 1)(2n)(2n - 1) by the same reasoning.

So you are given:
[(n+3)!] * [(2(n-1))!] / [n!] * [(2n+1)!]

=[(n+3)!] / [n!] * [(2(n-1))!] / [(2n+1)!]

=[(n + 3)(n + 2)(n + 1)] * [(2n - 2)!] / [(2n + 1)!]

=[(n + 3)(n + 2)(n + 1)] * [1 / [(2n + 1)(2n)(2n - 1)]
=[(n + 3)(n + 2)(n + 1)] / [(2n + 1)(2n)(2n - 1)]

Divide both numerator and denominator by "n^3"

=[(1 + 3/n)(1 + 2/n)(1 + 1/n)] / [(2 + 1/n)(2)(2 - 1/n)]
As n ---> infinity, "1/n" factors go to zero.
So,

lim {n ---> inf.} [(1 + 3/n)(1 + 2/n)(1 + 1/n)] / [(2 + 1/n)(2)(2 - 1/n)]
= [(1 + 0)(1 + 0)(1 + 0)] / [(2 + 0)(2)(2 - 0)]
= [(1)(1)(1)] / [(2)(2)(2)]
= 1/8

Answer 3

once you differentiate the numerator and the denominator ... evaluate the fraction on the minimize, the fee is the minimize of the unique fraction. it relatively is a rule of limits ... The differential of a factorial .. n(n-a million)(n-2)(n-3)...a million be conscious in case you differentiate this you get a chain that sounds like: (n-a million)(n-2)...a million + n(n-2)(n-3)...a million + n(n-a million)(n-3)...a million = n! [ a million/n + a million/(n-a million) + a million/(n-2) ... +a million] and because each and all the words a million/n, a million/(n-a million),... circulate to 0 whilst n ==> infinity different than the suitable term of +a million the differential of n! evaluated at n=infinity is n! it incredibly is the cost of the differentiated numerator ... n! The denominator differential is n2009^(n-a million) Now you're left with: n!/n2009^(n-a million) ... so which you do it lower back ... and you're left with n!/n(n-a million)2009^(n-2) and, as you proceed the approach you're left with n!/n!2009 which, once you evaluate it is going to become a million/2009^0 = a million consequently, minimize of n!/2009^n as n==>infinity = a million

Answer 4

You did not do it correctly. First, the expression can be written as:

n!(n+1)(n+2)(n+3)(2n-2)!
-----------------------------------
n!(2n-2)!(2n-1)(2n)(2n+1)

canceling factorials you are left with:

(n+1)(n+2)(n+3)
----------------------
(2n-1)(2n)(2n+1)

This approaches 1/8 as n -> infinity

Answer 5

2n^2 +4n-6
----------------
2n^2-6

When you apply the limit to your reduced form the answer is 1


If the answer 1 is wrong then the rduction could be wrong