用极限的运算法则就行. lim(x→0)2(x+1)=2 lim(x→0)arctan(1/x) =π/2 所以lim(x→0)2(x+1)arctan(1/x)=[lim(x→0)2(x+1)][lim(x→0)arctan(1/x)]=2×π/2=π 注：y=arctanx是一个增函数,x∈(-∞,+∞),y∈(-π/2,π/2)
