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Squeeze Thereom
Calculus I
Calculus II
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By Rick Stoll          

 Calc I

 

[Under Construction]

 

 

 

As a motivation let us consider the function

 

\begin{displaymath}f(x) = x^2 \sin\left(\frac{1}{x}\right)\;\cdot\end{displaymath}

 

When x get closer to 0, the function $\displaystyle g(x)=\sin\left(\frac{1}{x}\right)$ fails to have a limit. So we are not able to use the basic properties discussed in the previous pages. But we know that this function $\displaystyle g(x)=\sin\left(\frac{1}{x}\right)$ is bounded below by -1 and above by 1, i.e.

 

\begin{displaymath}-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\end{displaymath}

 

for any real number x. Since $x^2 \geq 0$, we get

 

\begin{displaymath}- x^2 \leq x^2\;\sin\left(\frac{1}{x}\right) \leq x^2 \;. \end{displaymath}

 

Hence when x get closer to 0, x2 and -x2 become very small in magnitude. Therefore any number in between will also be very small in magnitude. In other words, we have

 

\begin{displaymath}\lim_{x \rightarrow 0}\quad x^2\;\sin\left(\frac{1}{x}\right) = 0\;.\end{displaymath}

 

This is an example for the following general result:

Theorem: The "Squeeze" Theorem Assume that

 

\begin{displaymath}h(x) \leq f(x) \leq g(x)\end{displaymath}

 

for any x in an interval around the point a. If

 

\begin{displaymath}\lim_{x \rightarrow a} h(x) = L\;\; \mbox{and}\;\; \lim_{x \rightarrow a} g(x)= L\;,\end{displaymath}

 

then

 

\begin{displaymath}\lim_{x \rightarrow a} f(x) = L\;.\end{displaymath}

 

Example. Let f(x) be a function such that $\vert f(x)\vert \leq M$, for any $x \neq 0$. The Sandwich Theorem implies

 

\begin{displaymath}\lim_{x \rightarrow 0} x\;f(x) = 0\;.\end{displaymath}

 

Indeed, we have

 

\begin{displaymath}\vert x\; f(x)\vert \leq M \vert x\vert\end{displaymath}

 

which implies

 

\begin{displaymath}-M \vert x\vert \leq x\; f(x) \leq M \vert x\vert \end{displaymath}

 

for any $x \neq 0$. Since

 

\begin{displaymath}\lim_{x \rightarrow 0} M\;\vert x\vert = 0\;\;\mbox{and} \;\; \lim_{x \rightarrow 0} -M\;\vert x\vert = 0\;,\end{displaymath}

 

then the squeeze Theorem implies

 

\begin{displaymath}\lim_{x \rightarrow 0} x\;f(x) = 0\;.\end{displaymath}

 

 

Exercise 1. Use the squeeze Theorem to prove that

 

\begin{displaymath}\lim_{x \rightarrow a} \sqrt{x} = \sqrt{a}\end{displaymath}

 

for any a > 0.

 

 

 

 

Exercise 2. Use the squeeze Theorem to prove that

 

\begin{displaymath}\lim_{x \rightarrow 0} x \cos\left(\frac{1}{x}\right) = 0\;.\end{displaymath}

 

 

 

 

Exercise 3. Consider the function

 

\begin{displaymath}f(x) = \left\{ \begin{array}{lll} 1 + 2x^2 &\mbox{if $x$ is r... ...\\ 1 + x^4 &\mbox{if $x$ is irrational.}\\ \end{array}\right.\end{displaymath}

 

Use the squeeze Theorem to prove that

 

\begin{displaymath}\lim_{x \rightarrow 0} f(x) = 1\;.\end{displaymath}

 

 

 

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Last modified: April 30, 2001