夹逼法求极限

一、描述

使用夹逼法需要同时满足以下两个条件

• $ g(x) \leq f(x) \leq h(x)$
• $ \lim g(x) = \lim h(x) = A (A \epsilon R)$
  
  

因此可知，$ \lim f(x) = A$

二、Example

计算：

\[\lim_{n \rightarrow 0}({\frac{n}{n^2+1}}+{\frac{n}{n^2+2}}+\cdots +{\frac{n}{n^2+(n-1)}}+{\frac{n}{n^2+n}}) \]

解：令

\[f(n) = \frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots +\frac{n}{n^2+(n-1)}+\frac{n}{n^2+n} \]

\[g(n) = \frac{n}{n^2+1}+\frac{n}{n^2+1}+\cdots +\frac{n}{n^2+1}+\frac{n}{n^2+1} = \frac{n^2}{n^2+1} \]

\[h(n) = \frac{n}{n^2+n}+{\frac{n}{n^2+n}}+\cdots +\frac{n}{n^2+n}+\frac{n}{n^2+n} = \frac{n^2}{n^2+n} \]

因为：

\[ \lim_{n \rightarrow 0} g(n) = \lim_{n \rightarrow 0}\frac{n^2}{n^2+1} = 1 \]

\[ \lim_{n \rightarrow 0} h(n) = \lim_{n \rightarrow 0}\frac{n^2}{n^2+n} = 1 \]

根据夹逼定理：

\[ g(x) \leq f(x) \leq h(x) \]

\[ \lim g(x) = \lim h(x) = 1 \]

所以：

\[ \lim f(x) =1 \]