紧急:求lim tan(e^(x-1)-e^(x^2-1))/(arctanx-π/4),X趋于1;
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紧急:求lim tan(e^(x-1)-e^(x^2-1))/(arctanx-π/4),X趋于1;
![紧急:求lim tan(e^(x-1)-e^(x^2-1))/(arctanx-π/4),X趋于1;](/uploads/image/z/19065357-45-7.jpg?t=%E7%B4%A7%E6%80%A5%EF%BC%9A%E6%B1%82lim+tan%28e%5E%28x-1%29-e%5E%28x%5E2-1%29%29%2F%28arctanx-%CF%80%2F4%29%2CX%E8%B6%8B%E4%BA%8E1%EF%BC%9B)
利用等价无穷小和罗比达法则.
e^(x-1)-e^(x^2-1)趋于0,tanx与x等价.又因为分子,分母都趋于0,可以用罗比达法则.
如下
lim tan(e^(x-1)-e^(x^2-1))/(arctanx-π/4)=(e^(x-1)-e^(x^2-1))/(arctanx-π/4)=(e^(x-1)-2xe^(x^2-1))(1+x^2)
然后将x=1代入得
lim tan(e^(x-1)-e^(x^2-1))/(arctanx-π/4)=(e^(x-1)-e^(x^2-1))/(arctanx-π/4)=(e^(x-1)-2xe^(x^2-1))(1+x^2)=-2
e^(x-1)-e^(x^2-1)趋于0,tanx与x等价.又因为分子,分母都趋于0,可以用罗比达法则.
如下
lim tan(e^(x-1)-e^(x^2-1))/(arctanx-π/4)=(e^(x-1)-e^(x^2-1))/(arctanx-π/4)=(e^(x-1)-2xe^(x^2-1))(1+x^2)
然后将x=1代入得
lim tan(e^(x-1)-e^(x^2-1))/(arctanx-π/4)=(e^(x-1)-e^(x^2-1))/(arctanx-π/4)=(e^(x-1)-2xe^(x^2-1))(1+x^2)=-2
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