高数，求极限问题

(3)
y=-x
lim(x->-∞) ln(1+3^x)/ln(1+2^x)
=lim(y->+∞) ln(1+3^(-y))/ln(1+2^(-y))
=lim(y->+∞) ln[(1+3^y)/ 3^y ]/ln[ (1+2^y)/2^y ] (0/0 分子分母分别求导)
=lim(y->+∞) [ (ln3).3^y/(1+3^y) - ln3 ] /[ (ln2).2^y/(1+2^y) -ln2 ]
=lim(y->+∞) [ -ln3/(1+3^y) ] /[ -ln2/(1+2^y) ]
=(ln3/ln2) lim(y->+∞) (1+2^y)/(1+3^y)
=0
(4)
y=-x
lim(x->-∞) ln(1+e^x)/x
=lim(y->+∞) ln[1+e^(-y)]/(-y)
=-lim(y->+∞) ln[(1+e^y)/e^y ]/y (0/0 分子分母分别求导)
=-lim(y->+∞) [ e^y/(1+e^y) -1 ]
=lim(y->+∞) [ 1/(1+e^y) ]
=0
(5)
L=lim(x->π/2) (sinx)^tanx
lnL
=lim(x->π/2) ln(sinx)/ cotx (0/0 分子分母分别求导)
=lim(x->π/2) (cosx/sinx)/ [-(cscx)^2 ]
=0
=> L=1
lim(x->π/2) (sinx)^tanx = 1
(6)

lim(x->π/2) [(1+tanx)/(1+sinx) ]^(1/sinx) -> +∞
是不是这样
lim(x->0) [(1+tanx)/(1+sinx) ]^(1/sinx)
L =lim(x->0) [(1+tanx)/(1+sinx) ]^(1/sinx)
lnL
=lim(x->0) ln[(1+tanx)/(1+sinx) ]/sinx
=lim(x->0) ln[(1+tanx)/(1+sinx) ]/x (0/0 分子分母分别求导)
=lim(x->0) [(secx)^2/(1+tanx) - cosx/(1+sinx) ]
=1-1
=0
=> L=1
lim(x->0) [(1+tanx)/(1+sinx) ]^(1/sinx) =1

方法如下图所示，
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如图所示