求极限limx^2sin1/(x+x^2) (x趋向0)

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求极限limx^2sin1/(x+x^2)(x趋向0)求极限limx^2sin1/(x+x^2)(x趋向0)求极限limx^2sin1/(x+x^2)(x趋向0)如果是:limx^2sin【1/(x+

求极限limx^2sin1/(x+x^2) (x趋向0)
求极限limx^2sin1/(x+x^2) (x趋向0)

求极限limx^2sin1/(x+x^2) (x趋向0)
如果是:
limx^2sin【1/(x+x^2)】
答案为0
sin【1/(x+x^2)】是有界函数,x²是无穷小.有界函数×无穷小为无穷小
如果是:
lim(x^2sin1)/(x+x^2)
罗比达法则可得:
lim(2xsin1)/(1+2x)
=0