利用等价无穷小：ln(1+ax^2)~ax^2
sinx~x
lim[(1+ax^2)/x sinx=lim(1+ax^2)/limxsin=ax^2/x*x=a(x趋近于0负)
lim [ln(1+ax^2)]/(x sinx) (x趋近于0负）
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