If x 0 ( xx ) 1x2 = p, then 96 e p is equal to
Answer
32
Solution
The limit is of the form 1 , so
p = eL, where L = x 0 1x2 ( xx - 1 ) = x 0 x - xx3
∵ x = x + x33 + 2x515 + ,
L = x 0 ( 13 + 2x215 + ) = 13
so p = e1/3.
∴ 96 e p = 96 13 = 32
Answer: 32