Question

大一高数 请问这题怎么算呀？谢谢大家！

题目如图

Answer 1

三角函数变换1-cosx=2(sinx/2)^2
=limn^2*2sin((5.5/2n))^2
由两个重要极限
x趋向0,sinx~x
则原式=limn^2*2(5.5/2n)^2
=2*5.5^2/4
=15.125

Answer 2

1-cos(5.5/n) = 2(sin(5.5/2n))^2 ~ 2 *(5.5/2n)^2
答案应该是 2 * (5.5)^2/2^2