求极限六道题

(9)
let
y=1/x
lim(x->+∞) x[ln(x+1)-lnx ]
=lim(y->0) [ln(1+1/y)-ln(1/y) ] /y
=lim(y->0) ln(1+y) /y
=lim(y->0) y/y
=1
(10)
x->0
e^x ~ 1+x
lim(x->0)( e^x -1)/x
=lim(x->0)x/x
=1
(11)
lim(x->+∞) [( x-1)/x ]^x
=lim(x->+∞) [ 1-1/x ]^x
=e^(-1)
(12)
(2x+3)/(2x+1)
=1 + 2/(2x+1)
let
1/y=2/(2x+1)
lim(x->+∞) [(2x+3)/(2x+1)]^(x+1)
=lim(x->+∞) [1 + 2/(2x+1)]^(x+1)
=lim(y->+∞) [1 + 1/y]^[( 2y-2)/2 +1 ]
=lim(y->+∞) [1 + 1/y]^y
=e
(13)
lim(x->1) [√(3-x) -√(1+x) ]/(1-x^2)
=lim(x->1) [(3-x) -(1+x) ]/{ (1-x^2) .[√(3-x) +√(1+x) ] }
=lim(x->1) 2(1-x)/{ (1-x^2) .[√(3-x) +√(1+x) ] }
=lim(x->1) 2/{ (1+x) .[√(3-x) +√(1+x) ] }
=2/{ 2 .[√2 +√2 ] }
=√2/4
(14)
lim(x->1) [x^(1/3) -1]/(√x -1 ) (0/0)
分子，分母分别求导
=lim(x->1) (1/3)x^(-2/3) -1]/ [ (1/2)x^(-1/2) ]
=(2/3)lim(x->1) x^(-1/6)
=2/3