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Squeeze theorem in Calculus-1 course

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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 \[f(x)-g(x)\to 0\] \[f(x)=(f(x)-g(x))+g(x)\to 0+L=L.\]

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

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

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

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

Note by Yin Zhao
3 years, 1 month ago

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