x→∞,求[(3+x)/(6+x)]^[(x-1)/2]的极限

Question

x→∞,求[(3+x)/(6+x)]^[(x-1)/2]的极限

Answer 1

介绍两种方法
法一：(3+X)/(6+X)]^[(X-1)/2
=[1-3/(x+6)]^(x-1)/2
=[1-3/(x+6)]^[-(x+6)/3]*[(x-1)/2]*[-3/(x+6)]
所以原式=e^lim[(x-1)/2]*[-3/(x+6)]=e^(-3/2)
利用重要极限lim(x→0）(1+x)^(1/x)=e
法二：取对数求极限法
设y=[(3+X)/(6+X)]^[(X-1)/2]
则limlny=[(x-1)/2]ln[(x+3)/(x+6)]
=limln[(x+3)/(x+6)]/[2/(x-1)]
上式为不定式0/0型,使用洛必达法则
=lim[3/(x+3)(x+6)]/[(-2)/(x+1)^2]
=-3/2
所以limy=lime^lny=e^limlny=e^(-3/2)