一些标准的极限

前置

• converge

一些标准的极限

Proposition Let $0<x<1$. Then we have ${\lim }_{n\to \infty }{x}^n=0$.

Proof Since $0<x<1$, one can show that the sequence $(x^n{)}_{n=1}^{\infty }$ is decreasing (why?). On the other hand, the sequence $(x^n{)}_{n=1}^{\infty }$ has a lower bound of $0$. Thus the sequence $(x^n{)}_{n=1}^{\infty }$ converges to some limit $L$. Since $x^{n+1}=x\times x^n$, we thus see from the limit laws that $(x^{n+1}{)}_{n=1}^{\infty }$ converges to $xL$. But the sequence $(x^{n+1}{)}_{n=1}^{\infty }$ is just the sequence $(x^n{)}_{n=2}^{\infty }$ shifted by one, and so they must have the same limits (why?). So $xL=L$. Since $x\ne 1$, we can solve for $L$ to obtain $L=0$. Thus $(x^n{)}_{n=1}^{\infty }$ converges to $0$.

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Proposition 对于任何 $x>0$，均有：${\lim }_{n\to \infty }{x}^{1/n}=1$

证明：
过程我们分三种情况讨论：情况 1：当 $x=1$ 时如果 $x=1$，那么对于任何正整数 $n$，都有 $x^{1/n}=1^{1/n}=1$。显然：${\lim }_{n\to \infty }1=1$情况 2：当 $x>1$ 时令 $x^{1/n}=1+h_n$。由于 $x>1$，显然有 $h_n>0$。根据二项式定理（或更简单的伯努利不等式 $(1+h{)}^n\ge 1+nh$）：$x=(1+h_n{)}^n\ge 1+n{h}_n$由此可得：$0<h_n\le \frac{x-1}{n}$当 $n\to \infty$ 时，$\frac{x-1}{n}\to 0$。根据夹逼定理：${\lim }_{n\to \infty }{h}_n=0$因此：${\lim }_{n\to \infty }{x}^{1/n}={\lim }_{n\to \infty }(1+h_n)=1+0=1$情况 3：当 $0<x<1$ 时令 $y=\frac{1}{x}$。因为 $0<x<1$，所以 $y>1$。根据情况 2 的结论，我们知道：${\lim }_{n\to \infty }{y}^{1/n}=1$利用极限的商法则：${\lim }_{n\to \infty }{x}^{1/n}={\lim }_{n\to \infty }{\left(\frac{1}{y}\right)}^{1/n}={\lim }_{n\to \infty }\frac{1}{y^{1/n}}=\frac{1}{1}=1$结论综上所述，对于任何 $x>0$，均有：${\lim }_{n\to \infty }{x}^{1/n}=1$

Q.E.D.

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