Question

limx→π/2(ln(sinx))/(π-2x)^2求极限

Answer 1

lim(x→π/2) (ln(sinx))/(π-2x)^2 (0/0)
=lim(x→π/2) cotx/[-4(π-2x)] (0/0)
=lim(x→π/2) -(cscx)^2/8
=-1/8

追问

(π-2x)^2怎么算出-4(π-2x)？

追答

[(π-2x)^2]'

=2(π-2x)(-2)
=-4(π-2x)

追问

cotx=cscx^2?
书上是说1/sinx*cosx=1/sin^2x-csc^2x

追答

(lnsinx)'
= (1/sinx). (sinx)'
= cosx/sinx
=cotx

(cotx)' = -(cscx)^2

追问

cscπ/2=1?

追答

cscx= 1/sinx
csc(π/2) = 1/sin(π/2) = 1