Question

以e为底n的除n的极限为0怎么证明?

Answer 1

不难,有好几个：

证1 （用定义）记 n^(1/n) = 1+h[n],有 h[n]>0,且

n = (1+h[n])^n > C(n,2)(h[n])^2 = [n(n-1)/2](h[n])^2,

于是,有

0 < h[n] < √[2/(n-1)],

进而,有

lnn/n = lnn^(1/n) = ln(1+h[n]) < h[n] < √[2/(n-1)].

对任意ε>0,要使

|lnn/n| = lnn^(1/n) < √[2/(n-1)] < ε,

只需 n > 2/(ε^2)+1,取 N=[2/(ε^2)]+2,则当 n>N 时,有

|lnn/n| = lnn^(1/n) < √[2/(n-1)] < … < ε,

得证

lim(n→∞)lnn/n = 0.

证2 （用夹逼定理）记 n^(1/n) = 1+h[n],有 h[n]>0,且

n = (1+h[n])^n > C(n,2)(h[n])^2 = [n(n-1)/2](h[n])^2,

于是,有

h[n] < √[2/(n-1)],

进而,有

0 < lnn/n = lnn^(1/n) = ln(1+h[n])

< h[n] < √[2/(n-1)]

→0 (n→∞),

据夹逼定理,得知

lim(n→∞)lnn/n = 0.

证3 （利用 L'Hospital 法则）

lim(n→∞)lnn/n

= lim(x→+∞)lnx/x (∞/∞)

= lim(x→+∞)(1/x)/1

= 0.