((3x+1)/(3+x))^(1/(x-1)) 当x趋向于1时的极限

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((3x+1)/(3+x))^(1/(x-1)) 当x趋向于1时的极限

((3x+1)/(3+x))^(1/(x-1)) 当x趋向于1时的极限
((3x+1)/(3+x))^(1/(x-1)) 当x趋向于1时的极限

((3x+1)/(3+x))^(1/(x-1)) 当x趋向于1时的极限
设 f(x) = ((3x+1)/(3+x))^(1/(x-1))
ln f(x) = 1/(x-1) * ln[(3x+1) /(x+3) ] = 1/(x-1) * ln[ 1+ 2(x-1) /(x+3) ]
当x->1 时,2(x-1) /(x+3) ->0,ln[ 1+ 2(x-1) /(x+3) ] 2(x-1) /(x+3)
lim(x->1) lnf(x)
= lim(x->1) [2(x-1) /(x+3)] /(x-1) 等价无穷小代换
= lim(x->1) 2/(x+3) = 1/2
原式 = e^(1/2) = √e

e^(-2)
((3x+1)/(3+x))^(1/(x-1)) =e^(ln((3x+1)/(3+x))/(x-1));
运用罗比达法则，e的指数部分为{(3+x)/(3x+1)}*{(3x+1)-3*(x+3)}/(x+3)^2,消去了（x-1）项，
把x=1代入得到指数部分为-2
所以为e^(-2)

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