求lim(x^2sin(1/x^2)/sinx) x-->0用计算机算是0,但方法(1):原式=lim[sin(1/x^2)/(1/x^2)*x/sin(x)/x]=lim(1/x)=无穷大 方法(2):原式=lim[(2xsin(1/x^2)+
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求lim(x^2sin(1/x^2)/sinx) x-->0
用计算机算是0,但方法(1):原式=lim[sin(1/x^2)/(1/x^2)*x/sin(x)/x]=lim(1/x)=无穷大 方法(2):原式=lim[(2xsin(1/x^2)+x^2*cos(1/x^2)*(-2)/x^3)/cos(x)]=无穷大
为什么答案会不一样?
lim(x->0)(x^2sin(1/x^2)/sinx)
=lim(x->0)(x^2sin(1/x^2)/(x)
=lim(x->0)xsin(1/x^2)
因为
lim(x->0)x=0,即为无穷小
而|sin(1/x^2)|≤1,即为有界函数
由性质,无穷小和有界函数的乘积为无穷小,所以
原式=0
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