7. The following is a well-known result known as Stirling's formula: lim n! 1 TL " > ne "2n For this problem, you may assume we already know this identity to be true. Use it to calculate the limits of the four sequences below. n!e" n > n | 1/2 (a) lim (b) lim (2n)! 211 11 100 e 2n (2n) 2" n (c) lim 1L 100 (d) lim (2n)! n (n!)2 4n Vn! n 7. (a) 2 (b) 2 1 1 (c) (d) =-