求 lim(x→∞)[1-(3/x)]^x的极限.

lim(x→∞)[1-(3/x)]^x
=lim(x→∞)e^{ln[1-(3/x)]^x}
=e^lim(x→∞)xln[1-(3/x)]
lim(x→∞)xln[1-(3/x)]
=lim(x→∞){ln[1-(3/x)]}/(1/x)
由洛比达法则
=lim(x→∞)[x/(x-3)]*(-3/-x^2)/[1/(-x^2)]
=lim(x→∞)[x/(x-3)])]*-3
=-3
所以lim(x→∞)[1-(3/x)]^x=e^lim(x→∞)xln[1-(3/x)]=e^（-3）
本来很简单的题.在电脑上写出来看起来很复杂.Orz
用笔照这过程写一遍应该知道了.

lim[1-(3/x)]^x
=lim{[1+(-3/x)]^(-x/3)}^(-3)
因为lim[1+(1/t)]^t=e
=e^(-3)

lim(x→∞)[1-(3/x)]^x
=lim(x→∞)[1-(3/x)]^(3x/3)
=lim(x→∞){[1-(3/x)]^(x/3)}^3
=1/e^3
因为lim(x→∞){[1-(3/x)]^(x/3)=1/e

lim(x→0)[1-(3/x)]^x
代换x=-3t x→0 t→0
=lim(t→0)(1+1/t)^(-3t)
=lim(t→0)[(1+1/t)^t]^(-3)
=e^(-3)