Question

求一道求极限的高数题，lim（x趋近于无穷）[(x+1)/（x-1）]∧x

Answer 1

解法一：原式=lim(x->∞){[(1+2/(x-1))^((x-1)/2)]^[2x/(x-1)]}
={lim(x->∞)[(1+2/(x-1))^((x-1)/2)]}^{lim(x->∞)[2x/(x-1)]}
=e^{lim(x->∞)[2x/(x-1)]}
(应用重要极限lim(z->∞)[(1+1/z)^z]=e)
=e^{lim(x->∞)[2/(1-1/x)]}
=e^[2/(1-0)]
=e²；
解法二：∵lim(x->∞){xln[(x+1)/(x-1)]}=lim(x->∞){[ln(1+1/x)-ln(1-1/x)]]/(1/x)}
=lim(y->0){[ln(1+y)-ln(1-y)]]/y}
(令y=1/x)
=lim(y->0)[1/(1+y)+1/(1-y)]
(0/0型极限，应用罗比达法则)
=1/(1+0)+1/(1-0)
=2
∴原式=lim(x->∞){e^[xln((x+1)/(x-1))]
=e^{lim(x->∞)[xln((x+1)/(x-1))]}
=e^(2)
=e²。

Answer 2

为简便省去极限号
分子分母同时除以x,
原式=[(1+1/x)/(1-1/x)]^x=[(1+1/x)]^x/[(1-1/x)]^x=[(1+1/x)]^x乘以[(1-1/x)]^{-x}=e×e=e²