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 \[\lim_{x\to a}f(x)=L.\]


(1) When L=0L=0

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

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

Example1\boxed{Example 1} limn(1n+1+1n+2++1n+n)=?\lim_{n\to\infty}(\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n})=?

Example2\boxed{Example 2} limn(1n2+n+1+2n2+n+2++nn2+n+n)=?\lim_{n\to\infty}(\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n})=?

Solution1\boxed{Solution 1} Let Sn=1n+1+1n+2++1n+n, then\text{Let}\ S_n=\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n},\ \text{then} nn+n<Sn<nn+1\frac{n}{n+\sqrt n}<S_n<\frac{n}{n+1} Note that limnnn+n=limnnn+1=1\text{Note that}\ \lim_{n\to\infty}\frac{n}{n+\sqrt n}=\lim_{n\to\infty}\frac{n}{n+1}=1 limn(1n+1+1n+2++1n+n)=1\Rightarrow\lim_{n\to\infty}(\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n})=1

Solution2\boxed{Solution 2} Let Sn=1n2+n+1+2n2+n+2++nn2+n+n,\text{Let}\ S_n=\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n}, 1n2+n+n+2n2+n+n++nn2+n+n\Rightarrow \frac{1}{n^2+n+n}+\frac{2}{n^2+n+n}+\cdots +\frac{n}{n^2+n+n} <1n2+n+1+2n2+n+2++nn2+n+n<\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n} <1n2+n+1+2n2+n+1++nn2+n+1<\frac{1}{n^2+n+1}+\frac{2}{n^2+n+1}+\cdots +\frac{n}{n^2+n+1} 12n(n+1)n2+n+n<Sn<12n(n+1)n2+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} Note that limn12n(n+1)n2+n+n=limn12n(n+1)n2+n+1=12\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} limn(1n2+n+1+2n2+n+2++nn2+n+n)=12\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
7 years, 1 month ago

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