高中数学解法
lim(x→∝) (1-1/X)^X
=lim(x→∝) (1+1/(-X))^X
=lim(x→∝) 1/[(1+1/(-X))^(-X)]
=1/[lim(x→∝) (1+1/(-X))^(-X)]
=1/e
高等数学解法
lim(x→∝) (1-1/X)^X
=(令1/x=t)=lim(t→0) (1-t)^(1/t)
=lim(t→0) e^[ln(1-t)^(1/t)]
=e^lim(t→0) [ln(1-t)]/t
=洛必达法则=e^lim(t→0) -1/(1-t)
=e^[-1/(1-0)]
=e^(-1)
=1/e
lim(x→∞)(1-3/x)^x
=lim(x→∞)(1-3/x)^(-x/3)^(-3)
=e^(-3)
注:^——表示次方
lim [x→1] [1/(x-1) - 1/lnx]
通分
=lim [x→1] (lnx-x+1)/[(x-1)lnx]
lnx=ln(1+x-1)等价于x-1,分母的lnx换为x-1
=lim [x→1] (lnx-x+1)/(x-1)²
洛必达法则
=lim [x→1] (1/x - 1)/[2(x-1)]
=lim [x→1] (1-x)/[2x(x-1)]
=lim [x→1] -1/(2x)
=-1/2
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lim[(2-1/x)^30*(3-2/x)^20]/(1+1/x)^50
=2^30*3^20
分子分母同除以X^50
lim (x-1)/(x+1)
x→∞
=lim (1- 1/x)/(1+ 1/x)
x→∞
=(1-0)/(1+0)
=1
上下除以x
原式=lim1/[√(2-1/x²)
=1/√(2-0)
=√2/2
极限是0,用洛必达法则上下求导一次就出来了
求极限 x→∞lim{x²[1-xsin(1/x)]}
解:令1/x=u,则x=1/u, x→∞时u→0.
故原式=u→0lim{[1-(1/u)sinu]/u²}=u→0lim[(u-sinu)/u³]
=u→0lim[(1-cosu)/(3u²)]=u→0lim[(sinu)/6u]=u→0lim[(cosu)/6]=1/6.
x→∞时,
[(x-1)/(x+1)]^(3x)
={[1-2/(x+1)]^3(x+1)/2-3/2}^2
→(1/e^3)^2
=1/e^6.
lim趋于无穷 x^2-1/2x^2-x-1
=lim趋于无穷 (x+1)(x-1)/(x-1)(2x+1)
==lim趋于无穷 (x+1)/(2x+1)
=1/2