几道求极限问题lin (x^3-3x+2)/(x-1)^2=x->+∞lin [1/x(sinx)+xsin(1/x)]
1个回答

用罗必塔法则啊 很简单

对0/0,∞/∞,先对分子分母分别求导,至可以求出极限

lin (x^3-3x+2)/(x-1)^2= lin(3x^2-3)/2(x-1)=lin(6x)/2=+∞

x->+∞ x->+∞ x->+∞

lin 1/x(sinx)+xsin(1/x)]= lin 1/x(sinx)+lin xsin(1/x)

x->+∞ x->+∞ x->+∞

=0+lin((sin(1/x))/(1/x))=0+1=1

x->+∞

因为|1/x(sinx）|=0,故前者极限为0,

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