告诉极限值,求常数k ,

=lim(x->∞)[1+3k/(x-2k)]^x
=e^lim(x->∞)3kx/(x-2k)
=e^lim(x->∞)3k/(1-2k/x)
=e^3k
所以有 e^3k=8
两边取对数得 3k=ln8
k=(ln8)/3
再问： 这个是怎么得来的额=lim(x->∞)[1+3k/(x-2k)]^x 题目不是=lim(x->∞)[(x+k)/(x-2k)]^x吗
再答： 这么得来的： lim(x->∞)[(x+k)/(x-2k)]^x =lim(x->∞){[(x-2k)+3k]/(x-2k)}^x =lim(x->∞){[(x-2k)/(x-2k)+3k/(x-2k)]}^x =lim(x->∞)[1+3k/(x-2k)]^x