微积分求极限 lim(x->0+)sin(arct - 知识问答

微积分求极限 lim(x->0+)sin(arctan1/x) lim(x->π/4)【（cos2x）的立方】

3个回答

lim(x->0+)sin(arctan1/x)
=lim(ctgt->0+)sint,令arctan1/x=t得到.
=lim(t->π/2)sint
=1.
lim(x->π/4)（cos2x）^3
=cos(π/2)^3 ,直接代入.
=0

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