怎样理解函数y= cosx在x=0处的单调性

Question

如题所述

Answer 1

lim(x->0) [√(1+tanx) -√(1+sinx) ]/( x√[1+(sinx)^2] -x )
=lim(x->0) (tanx-sinx)/ [x(sinx)^2] (0/0)
=lim(x->0) [(secx)^2-cosx]/[xsin(2x) + (sinx)^2] (0/0)
=lim(x->0) [2(secx)^2.tanx+sinx]/[2xcos(2x)+2sin(2x)] (0/0)
=(1/2)lim(x->0) [2(secx)^4+4(secx)^2.(tanx)^2+cosx]/[-2xsin(2x) + 3cos(2x)]
=(1/2)lim(x->0) [2(secx)^4+4(secx)^2.(tanx)^2+cosx]/[-2xsin(2x) + 3cos(2x)]
=1/2