LIMx→0+ (sinx) ^x的极限用罗比达法则求极限,
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![LIMx→0+ (sinx) ^x的极限用罗比达法则求极限,](/uploads/image/z/4095133-61-3.jpg?t=LIMx%E2%86%920%2B+%28sinx%29+%5Ex%E7%9A%84%E6%9E%81%E9%99%90%E7%94%A8%E7%BD%97%E6%AF%94%E8%BE%BE%E6%B3%95%E5%88%99%E6%B1%82%E6%9E%81%E9%99%90%2C)
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,
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