LIMx→0+ (sinx) ^x的极限用罗比达法则求极限,

LIMx→0+ (sinx) ^x的极限
用罗比达法则求极限,

取对数
ln (sinx) ^x
=xlnsinx
=lnsinx/ (1/x)
罗比达法则
= cosx/sinx /(-1/x²)
= -x²cosx/sinx
=【-2xcosx+x²sinx】/cosx
=0
所以原始还原
=e^0
=1