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limx趋向于无穷(1÷x^2)ArCtAn[(x+1+x^2)/(x+1)(x%...

lim(x->∞) (1/x^2)arctan{ (x^2+x+1)/[(x+1)(x-2)] } =lim(x->∞) (1/x^2) .lim(x->∞) arctan{ (x^2+x+1)/[(x+1)(x-2)] } =lim(x->∞) (1/x^2) .lim(x->∞) arctan{ (1+1/x+1/x^2)/[(1+1/x)(1-2/x)] } =lim(x->∞) (1/x^2) . (π/4) =0

如图

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无穷小替换,arctanx=x =[ax-ax/(1+x)]/x^2 =[ax+ax^2-ax]/(1+x)/x^2=a/(1+x)=a

limx趋向无穷(1+-1/1+x)^-1/2=1

如图所示:渐近线分别是x = 0和y = π/4

原式 =lim((√x^2+1)+(√x^2-1))((√x^2+1)-(√x^2-1))/((√x^2+1)+(√x^2-1)) =lim(x^2+1-(x^2-1))/((√x^2+1)+(√x^2-1)) =lim2/((√x^2+1)+(√x^2-1)) =0

1的无穷大次方型的,可以用这个公式: lim u^v =lim e^ (v(u-1)) (证明: lim u^v =lim e^ (vlnu)=lim e^ (v ln(1+u-1))=lim e^[v(u-1)] ,最后一步用到等价无穷小ln(1+x)~x ) 可以直接用那个公式,或者依照证明的那个思路解。

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