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limx→π/2(ln(sinx))/(π-2x)^2求极限
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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
=lim(x→π/2) cotx/[-4(π-2x)] (0/0)
=lim(x→π/2) -(cscx)^2/8
=-1/8
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追问
(π-2x)^2怎么算出-4(π-2x)?
追答
[(π-2x)^2]'
=2(π-2x)(-2)
=-4(π-2x)
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