Squeeze theorem in Calculus-1 course

Squeeze theorem

Assume that functions \(f,\ g,\ h\) satisfy \[g(x)\le f(x)\le h(x)\] and \[\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L.\] Then \[\lim_{x\to a}f(x)=L.\]

$\boxed{\textbf{Proof}}$

(1) When $L=0$

Assuming that $\forall x\in\textbf{R},\ g(x)=0,\ \lim_{x\to a}h(x)=0$ We have $\forall\varepsilon>0,\ \exists\delta>0:\ 0<|x-a|<\delta\Rightarrow |h(x)|<\varepsilon$ Because $0=g(x)\le f(x)\le h(x)$, so $|f(x)|\le|h(x)|$, which means $\forall\varepsilon>0,\ \exists\delta>0:\ 0<|x-a|<\delta\Rightarrow |f(x)|\le|h(x)|<\varepsilon$ $\Rightarrow\lim_{x\to a}f(x)=0.$

(2) General situation $g(x)\le f(x)\le h(x)\Rightarrow 0\le f(x)-g(x)\le h(x)-g(x)$ When $x\to a$ $h(x)-g(x)=L-L=0$ According to (1), we have $f(x)-g(x)\to 0$ $f(x)=(f(x)-g(x))+g(x)\to 0+L=L.$

$\boxed{Example 1}$ $\lim_{n\to\infty}(\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n})=?$

$\boxed{Example 2}$ $\lim_{n\to\infty}(\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n})=?$

$\boxed{Solution 1}$ $\text{Let}\ S_n=\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n},\ \text{then}$ $\frac{n}{n+\sqrt n}<S_n<\frac{n}{n+1}$ $\text{Note that}\ \lim_{n\to\infty}\frac{n}{n+\sqrt n}=\lim_{n\to\infty}\frac{n}{n+1}=1$ $\Rightarrow\lim_{n\to\infty}(\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n})=1$

$\boxed{Solution 2}$ $\text{Let}\ S_n=\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n},$ $\Rightarrow \frac{1}{n^2+n+n}+\frac{2}{n^2+n+n}+\cdots +\frac{n}{n^2+n+n}$ $<\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n}$ $<\frac{1}{n^2+n+1}+\frac{2}{n^2+n+1}+\cdots +\frac{n}{n^2+n+1}$ $\Rightarrow \frac{\frac{1}{2}n(n+1)}{n^2+n+n}<S_n<\frac{\frac{1}{2}n(n+1)}{n^2+n+1}$ $\text{Note that}\ \lim_{n\to\infty}\frac{\frac{1}{2}n(n+1)}{n^2+n+n}=\lim_{n\to\infty}\frac{\frac{1}{2}n(n+1)}{n^2+n+1}=\frac{1}{2}$ $\Rightarrow\lim_{n\to\infty}(\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n})=\frac{1}{2}$