怎么计算lim(x->0+)x^(1/2)*ln(sinx)?

怎么计算lim(x->0+)x^(1/2)*ln(sinx)?
怎么计算lim(x->0+)x^(1/2)*ln(sinx)?

怎么计算lim(x->0+)x^(1/2)*ln(sinx)?
0*∞的不定型,先化成比值,然后洛必达
ln(sinx)
= -------------------
x^(-1/2)
无穷/无穷
洛必达
(1/sinx)*cosx
= -------------------
(-1/2)x^(-3/2)
-2x^(3/2)cosx
= -------------------
sinx
x
= ------------ * [-2x^(1/2)cosx]
sinx
第一个的极限是1,方括号内的极限是0*1=0
所以原极限为1*0=0

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