当x趋于无穷时，【(x+3)/(x-1)】^x的极限？

Question

如题所述

Answer 1

lim(x->无穷大)[(x+3)/(x-1)]^x
对上式取对数
lim(x->无穷大)xln[(x+3)/(x-1)]=xln(x+3)-xln(x-1)=[ln(x+3)-ln(x-1)]/1/x 0/0型用罗必塔法则
=lim(x->无穷大)[1/(x+3)-1/(x-1)]/(-1/x^2)
=lim(x->无穷大)x^2/[1/(x-1)-1/(x+3)]
=lim(x->无穷大)x^2(x+3-x+1)/(x-1)(x+3)
=lim(x->无穷大)4x^2/(x-1)(x+3)
=4
所以lim(x->无穷大)[(x+3)/(x-1)]^x=e^4

Answer 2

lim(x->∞) [ 1 + 4/(x-1) ] ^ x
= lim(x->∞) [ 1 + 4/(x-1) ] ^ [ (x-1)/4 * 4x /(x-1) ]
= e^ 4
利用重要极限：lim(u->0) (1+u) ^(1/u) = e追问

第二步以后该怎么做呢？就是次方项那里怎么化简？

追答

lim(x->∞) [ 1 + 4/(x-1) ] ^ [ (x-1)/4] = e
幂指数 lim(x->∞) 4x / (x-1) = 4
∴ 原式 = e^4

本回答被提问者采纳

Answer 3

【(x+3)/(x-1)】^x=[1+4/(x-1)]^x=[1+4/(x-1)]^4{[(x-1)/4]}x[1+4/(x-1)]={[1+4/(x-1)]^[(x-1)/4]}^4=e^4