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)
求极限:lim 2(x+1)arctan(1/x) (x趋于0 )
求极限:lim 2(x+1)arctan(1/x) (x趋于0 )
数学人气:333 ℃时间:2019-09-27 06:52:37
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