Question

大一数学分析

Answer 1

x->0
sinx~ x- (1/6)x^3
sinx/x ~ 1-(1/6)x^2
--------------
1-cosx ~ (1/2)x^2
---------
lim(x->0) (sinx/x )^[1/(1-cosx)]
=lim(x->0) (1- (1/6)x^2 )^(2/x^2 )
=e^(-1/3)
(2)
L = lim(x->+∞) (lnx)^(1/x)
lnL
= lim(x->+∞) ln(lnx)/x (∞/∞)
= lim(x->+∞) 1/[x(lnx)]
=0
L=1
lim(x->+∞) (lnx)^(1/x) =1
(3)
lim(n->∞) [ n - 1/(e^(1/n) -1 ) ]
consider
lim(x->∞) [ x - 1/(e^(1/x) -1 ) ]
let
y=1/x
y->0
分子
e^y ~ 1+y + (1/2)y^2
e^y -1 -y ~ (1/2)y^2
分母
y .(e^y -1 ) ~ y^2
-----------
lim(x->∞) [ x - 1/(e^(1/x) -1 ) ]
=lim(y->0) [ 1/y - 1/(e^y -1 ) ]
=lim(y->0) [ e^y -1 - y ]/ [y .(e^y -1 ) ]
=lim(y->0) (1/2)y^2/ y^2
=1/2
=>
lim(n->∞) [ n - 1/(e^(1/n) -1 ) ] =1/2

Answer 2

这不懂！