sinxlnx的极限怎么求X→0+

Question

sinxlnx的极限怎么求 X→0+

Answer 1

lim[x→0+](x^sinx)
=lim[x→0+](sinxlnx)
=(lim[x→0+]((sinx/x)*(xlnx))
(lim(x->0+)sinx/x=1 )
=lim[x→0+](lnx/(1/x))
=lim[x→0+]((1/x)/(-1/x^2))(洛比塔法则)
=lim(x->0+)-x
=0