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