Question

请问微积分这题怎么做？

第二小题

Answer 1

这题用到了重要极限lim（x→∞）（1+1/x）^x=e
第二小题中lim[（x-2）/(x+2)]^2x=lim[1-4/(x+2)]^-(x+2)/4·-4/（x+2）·2x=e^(-8)

Answer 2

(2)

(x-2)/(x+2) = 1 - 4/(x+2)

let
1/y = 4/(x+2)
lim(x->∞) [(x-2)/(x+2)]^(2x)
=lim(x->∞) [1- 4/(x+2)]^(2x)
=lim(y->∞) (1- 1/y)^[2(4y-2)]
=lim(y->∞) (1- 1/y)^(8y)
=e^(-8)
(4)
lim(x->∞) ( 1- 1/x)^(√x)

=lim(x->∞) e^[ ln( 1- 1/x)/ (1/√x) ] (0/0分子分母分别求导)

=lim(x->∞) e^{ [1/(x-1) - 1/x] / [-(1/2)(1/x^(3/2)] }
=lim(x->∞) e^{ -2x^(3/2)/[x(x-1)] }
=e^0
=1
(6)
x->0
cosx ~ 1- (1/2)x^2
lim(x->0) (cosx)^[1/(1-cosx) ]
=lim(x->0) (1- (1/2)x^2)^{ 1/[(1/2)x^2] }
= e^(-1)

Answer 3

侠骨柔肠改编自原著《侠女》篇。