Question

高数limx→π/2 (sinx^(2/（π-2x））-1）/（π/2 -x）

求详细过程

Answer 1

x→π/2 -2<=(sinx^(2/(π-2x))-1<=0
极限不存在

追问

这个答案不对啊是1/2

追答

左右趋近的极限不一样

题目没问题？

追问

题目没问题，你题目写错啦，你看我发的那个图片，是sinx

追答

应这样写 limx→π/2 ((sinx)^(2/(π-2x)-1）/（π/2 -x）

Answer 2

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

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