Question

利用洛必达法则求极限，如图所示

Answer 1

Answer 2

(6)

lim(x->1) ( 2/lnx - x/ln√x )
=lim(x->1) ( 2/lnx - 2x/lnx )
=2lim(x->1) (1-x)/lnx (0/0分子分母分别求导）
=2lim(x->1) -1/(1/x)
=2lim(x->1) -x
=-2
(8)
let
u = π/4 -x
lim(x->π/4) ( tanx)^[tan(2x) ]

=lim(u->0) [ tan(π/4 -u) ]^[tan(π/2 -2u) ]
=lim(u->0) [tan(π/4 -u)]^[1/tan(2u) ]
=lim(u->0) [(1-tanu)/(1+tanu) ]^[1/tan(2u) ]
=lim(u->0) [ -1 + 2/(1+tanu) ]^[1/tan(2u) ]
=lim(u->0) [ 1 - 2/(1+u) ]^[1/(2u) ]
= e^(-1)