Squeeze theorem in Calculus-1 course

This note has been used to help create the Squeeze Theorem wiki

Squeeze theorem

Assume that functions f, g, hf,\ g,\ h satisfy g(x)f(x)h(x)g(x)\le f(x)\le h(x) and limxag(x)=limxah(x)=L.\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L. Then limxaf(x)=L.\lim_{x\to a}f(x)=L.

Proof\boxed{\textbf{Proof}}

(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
5 years, 2 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...