Find this limit as x approaches infinity?

lim_{xrarr \infty} (""^(2n)C_(n))^(1/n)

1 Answer
May 7, 2017

lim_{n rarr oo} (""^(2n)C_(n))^(1/n) = 4

Explanation:

I presume the question is meant to read n rarr oo, ie

L = lim_{n rarr oo} (""^(2n)C_(n))^(1/n)

Using the Combination Definition:

""^(n)C_r = (n!)/((n-r)!r!)

We can write:

L = lim_{n rarr oo} (((2n)!)/((2n-n)!n!))^(1/n)
\ \ = lim_{n rarr oo} (((2n)!)/(n!n!))^(1/n)
\ \ = lim_{n rarr oo} (((2n)!)/((n!)^2))^(1/n)

For large n we can approximate n! using Stirling's approximation formula:

n! ~~ sqrt(2 pi n) \ (n/e)^n

Which gives us:

L ~~ lim_{n rarr oo} ((sqrt(2 pi 2n) \ ((2n)/e)^(2n))/((sqrt(2 pi n) \ (n/e)^n)^2))^(1/n)

\ \ ~~ lim_{n rarr oo} ((sqrt(4 pi n) \ ((2n)/e)^(2n))/(2 pi n \ (n/e)^(2n)))^(1/n)

\ \ ~~ lim_{n rarr oo} ( (2sqrt(pi n)) /(2 pi n ) * (((2n)/e) / (n/e)) ^(2n))^(1/n)

\ \ ~~ lim_{n rarr oo} ( (2sqrt(pi n)) /(2 pi n ) * (2) ^(2n))^(1/n)

\ \ ~~ lim_{n rarr oo} ( 1/sqrt(pi n) )^(1/n) * lim_{n rarr oo}(4^n)^(1/n)

\ \ ~~ L_1 \ L_2 , say

If we look at the first limit:

L_1 = lim_{n rarr oo} ( 1/sqrt(pi n) )^(1/n)
\ \ \ \ = lim_{n rarr oo} ( (pi n)^(-1/2) )^(1/n)
\ \ \ \ = lim_{n rarr oo} (pi n)^(-1/(2n))
\ \ \ \ = 1

And the second limit:

L_2 = lim_{n rarr oo}(4^n)^(1/n)
\ \ \ \ = lim_{n rarr oo}4 = 4

Combining these results we get:

L ~~ 4

The approximation from Stirling's approximation formula becomes increasingly access as n increase, and as n rarr oo, we have:

L = lim_{n rarr oo} (""^(2n)C_(n))^(1/n) = 4