Difference between revisions of "Squeeze Theorem"

Revision as of 03:27, 6 May 2008

This is an AoPSWiki Word of the Week for May 4-11

The Squeeze Theorem (also called the Sandwich Theorem or the Squeeze Play Theorem) is a relatively simple theorem that deals with calculus, specifically limits.

Squeeze Theorem

Theorem

Suppose f(x) is between g(x) and h(x) for all x in the neighborhood of S. If g and h approach some common limit L as x approaches S, then \lim_{x\to S}f(x)=L.

Proof

If $f(x)$ is between $g(x)$ and $h(x)$ for all $x$ in the neighborhood of $S$ , then either $g(x)\leq f(x) \leq h(x)$ or $h(x)\leq f(x)\leq g(x)$ for all $x$ in the neighborhood of $S$ . Since the second case is basically the first case, we just need to prove the first case.

For all $\varepsilon >0$ , we must prove that there is some $\delta > 0$ for which $|x-S|<\delta \Rightarrow |f(x)-L|<\varepsilon$ .

Now since, $\lim_{x\to S}g(x)=\lim_{x\to S}h(x)=L$ , there must exist $\delta_1,\delta_2>0$ such that,

$|x-S|<\delta_1 \Rightarrow |g(x)-L|<\varepsilon$ and,

$|x-S|<\delta_2 \Rightarrow |h(x)-L|<\varepsilon.$

Now let $\delta = \min\{\delta_1,\delta_2\}$ . If $|x-S|<\delta$ , then

$-\varepsilon < g(x) - L \leq f(x) - L \leq h(x) - L < \varepsilon.$

So $|f(x)-L|<\varepsilon$ . Now by the definition of a limit, we get $\lim_{x\to S}f(x)=L$ .

@@ Line 5: / Line 5: @@
 ==Theorem==
-Suppose <math>f(x)</math> is between <math>g(x)</math> and <math>h(x)</math> for all <math>x</math> in the neighborhood of <math>S</math>. If <math>g</math> and <math>h</math> approach some common limit L as <math>x</math> approaches <math>S</math>, then <math>\lim_{x\to S}f(x)=L</math>.
+Suppose f(x) is between g(x) and h(x) for all x in the neighborhood of S. If g and h approach some common limit L as x approaches S, then \lim_{x\to S}f(x)=L.
 ==Proof==

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Difference between revisions of "Squeeze Theorem"

Revision as of 03:27, 6 May 2008

Theorem

Proof

See Also