l计算题已知lim趋近于无穷((x+c)/(x-c))^(x/2)=3,求c.,看了别人的解答,我有一部分不是很清楚lim(x→∞)[(x+c)/(x-c)]^(x/2)=lim(x→∞)[1+2c/(x-c)]^(x/2)=lim(x→∞)[1+2c/(x-c)]^{[(x-c)/(2c)]*[(cx)/(x-c)]}=e^lim(x→∞)

Question

l计算题 已知lim 趋近于无穷 ((x+c)/(x-c))^(x/2)=3,求c.,看了别人的解答,我有一部分不是很清楚
lim(x→∞) [(x+c)/(x-c)]^(x/2)
=lim(x→∞) [1+2c/(x-c)]^(x/2)
=lim(x→∞) [1+2c/(x-c)]^{[(x-c)/(2c)]*[(cx)/(x-c)]}
=e^lim(x→∞) [(cx)/(x-c)]
=e^c
=3
∴c=ln3
就是
=lim(x→∞) [1+2c/(x-c)]^{[(x-c)/(2c)]*[(cx)/(x-c)]}
=e^lim(x→∞) [(cx)/(x-c)]
这一步,我不懂,为很么e会直接出去了而且变成e^lim这样的形式,

Answer 1

lim(x→∞) [1+2c/(x-c)]^{[(x-c)/(2c)]*[(cx)/(x-c)]}从这步开始写
要知道一个极限式即lim(x→∞)(1+1/x)^x=e
lim(x→∞) [1+2c/(x-c)]^{[(x-c)/(2c)]*[(cx)/(x-c)]}这之中lim(x→∞) [1+2c/(x-c)]^[(x-c)/(2c)]=e
所以lim(x→∞) [1+2c/(x-c)]^{[(x-c)/(2c)]*[(cx)/(x-c)]}=lim(x→∞) e^[(cx)/(x-c)]=e^lim(x→∞) [(cx)/(x-c)]