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let
y=π/2-x
y->0
(cosy)^(1/y)
=e^[ln(cosy) /y ]
=e^{[ln[ (1 - (1/2)y^2 +o(y^2) ] /y }
=e^{ [ -(1/2)y^2 +o(y^2) ]/y }
=e^[ -(1/2)y + o(y) ]
=1 -(1/2)y + o(y)
(cosy)^(1/y) -1 = -(1/2)y + o(y)
lim(x->π/2) { (sinx)^[2/(π-2x)] -1 }/( π/2 - x)
=lim(y->0) [ (cosy)^(1/y) -1 ]/ y
=lim(y->0) -(1/2)y/ y
=-1/2
y=π/2-x
y->0
(cosy)^(1/y)
=e^[ln(cosy) /y ]
=e^{[ln[ (1 - (1/2)y^2 +o(y^2) ] /y }
=e^{ [ -(1/2)y^2 +o(y^2) ]/y }
=e^[ -(1/2)y + o(y) ]
=1 -(1/2)y + o(y)
(cosy)^(1/y) -1 = -(1/2)y + o(y)
lim(x->π/2) { (sinx)^[2/(π-2x)] -1 }/( π/2 - x)
=lim(y->0) [ (cosy)^(1/y) -1 ]/ y
=lim(y->0) -(1/2)y/ y
=-1/2
追问
答案是正1/2啊
追答
查一下问题有没有错
lim(x->π/2) { (sinx)^[2/(π-2x)] -1 }/( π/2 - x)
如果没有的话, 我肯定
lim(x->π/2) { (sinx)^[2/(π-2x)] -1 }/( π/2 - x) =-1/2
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