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1.求极限Iim x→0 sin2x/sin3x
2.求极限 Iim x→1 x^3-3x^2+2x/x^2-1
人气:188 ℃ 时间:2020-05-02 14:18:34
解答
limsin2x/sin3x
=lim(2/3)(sin2x/2x)(3x/sin3x)
=2/3
lim(x^3-3x^2+2x)/(x^2-1)
=lim(3x^2-6x+2)/2x (洛毕达法则)
=-1/2
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