Question

求数学极限 这两道题。详细过程，拜托了！





Answer 1

lim(n->∞) [ 2^(1/n) - 2^(1/(n+1) ) ]
= ( 2^0 -2^0 )
=0

lim(x->∞) [ sin(2/x)+cos(1/x) ]^x
let
y=1/x
lim(x->∞) [ sin(2/x)+cos(1/x) ]^x
=lim(y->0) [ sin(2y)+cosy ]^(1/y)
x->0
sin2x +cosy ~ 1+2y

lim(x->∞) [ sin(2/x)+cos(1/x) ]^x
=lim(y->0) [ sin(2y)+cosy ]^(1/y)
=lim(y->0) ( 1+ 2y )^(1/y)
=lim(z->0) ( 1+ z )^(2/z) ( z=2y)
=e^2

追问

第一个是38题，第二个是40题。你写的思路我也看得懂，只是跟答案的方法相差有点远，对答案的思路我有点不太明白。可以讲解一下吗

追答

那好！
n->∞
1/n ->0 , 1/(n+1) ->0
lim(n->∞) [ 2^(1/n) - 2^(1/(n+1) ) ]
= ( 2^0 -2^0 )
=0

(40)
等价无穷小
y->0
sin2y ~ 2y
cosy ~ 1-(1/2)y^2 ~ 1

lim(x->∞) [ sin(2/x)+cos(1/x) ]^x
=lim(y->0) [ sin(2y)+cosy ]^(1/y)
=lim(y->0) ( 1+ 2y )^(1/y)
=lim(z->0) ( 1+ z )^(2/z) ( z=2y)
=e^2