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计算极限lim(x→0)[1-x^2-e^(-x^2)]/(sin2x)^4
人气:221 ℃ 时间:2020-04-23 02:59:22
解答
lim(x→0)[1-x^2-e^(-x^2)]/(sin2x)^4 (等价无穷小代换)
=lim(x→0)[1-x^2-e^(-x^2)]/(16x^4) (0/0型 ,上下求导得)
=lim(x→0)[-2x+2xe^(-x^2)]/(32x^3)
=lim(x→0)[-1+e^(-x^2)]/(16x^2) (等价无穷小代换)
=lim(x→0)(-x^2)/(16x^2)
=-1/16第三步应该是lim(x→0)[-2x+2xe^(-x^2)]/(64x^3)吧?噢,脑筋短路了,应该乘4的。lim(x→0)[1-x^2-e^(-x^2)]/(sin2x)^4(等价无穷小代换)=lim(x→0)[1-x^2-e^(-x^2)]/(16x^4) (0/0型 ,上下求导得)=lim(x→0)[-2x+2xe^(-x^2)]/(64x^3)=lim(x→0)[-1+e^(-x^2)]/(32x^2)(等价无穷小代换)=lim(x→0)(-x^2)/(32x^2)=-1/32
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