Question

数学问题：用洛必达法则求极限。第5题。求教。

Answer 1

如图

Answer 2

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

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我需要时间

老师 您能写下来吗

太乱

追答

不是很乱
lim(x->π/2) ln(sinx)/(π- 2x)^2 (0/0)
d/dx[ ln(sinx )] = cotx ; d/dx [(π- 2x)^2] = -4(π- 2x)
=lim(x->π/2) cotx/ [-4(π- 2x)] (0/0)
d/dx (cotx) = -(cscx)^2 ; d/dx {-4(π- 2x)} = 8
=lim(x->π/2) -(cscx)^2/ 8
=-1/8

Answer 3

达法则求极限

追问

闭嘴