Find this limit as x approaches infinity?
#lim_{xrarr \infty} (""^(2n)C_(n))^(1/n)#
1 Answer
# lim_{n rarr oo} (""^(2n)C_(n))^(1/n) = 4#
Explanation:
I presume the question is meant to read
# 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! ~~ 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
# L = lim_{n rarr oo} (""^(2n)C_(n))^(1/n) = 4#