limx→π/2(ln(sinx))/(π-2x)^2求极限
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limx→π/2(ln(sinx))/(π-2x)^2求极限
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
再问: (π-2x)^2怎么算出-4(π-2x)?
再答: [(π-2x)^2]'
=2(π-2x)(-2)
=-4(π-2x)
再问: cotx=cscx^2?
书上是说1/sinx*cosx=1/sin^2x-csc^2x
再答: (lnsinx)'
= (1/sinx). (sinx)'
= cosx/sinx
=cotx
(cotx)' = -(cscx)^2
再问: cscπ/2=1?
再答: cscx= 1/sinx
csc(π/2) = 1/sin(π/2) = 1
=lim(x→π/2) cotx/[-4(π-2x)] (0/0)
=lim(x→π/2) -(cscx)^2/8
=-1/8
再问: (π-2x)^2怎么算出-4(π-2x)?
再答: [(π-2x)^2]'
=2(π-2x)(-2)
=-4(π-2x)
再问: cotx=cscx^2?
书上是说1/sinx*cosx=1/sin^2x-csc^2x
再答: (lnsinx)'
= (1/sinx). (sinx)'
= cosx/sinx
=cotx
(cotx)' = -(cscx)^2
再问: cscπ/2=1?
再答: cscx= 1/sinx
csc(π/2) = 1/sin(π/2) = 1
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