Difference between revisions of "Squeeze Theorem"

(There's no need for the inequalities to be strict)
(The old proof implicitly assumed that f converged. Replaced this with an epsilon-delta proof.)
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If <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>, then either <math>g(x)\leq f(x) \leq h(x)</math> or <math>h(x)\leq f(x)\leq g(x)</math> for all <math>x</math> in the neighborhood of <math>S</math>. Since the second case is basically the first case, we just need to prove the first case.
 
If <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>, then either <math>g(x)\leq f(x) \leq h(x)</math> or <math>h(x)\leq f(x)\leq g(x)</math> for all <math>x</math> in the neighborhood of <math>S</math>. Since the second case is basically the first case, we just need to prove the first case.
  
If <math>g(x)</math> increases to <math>L</math>, then <math>f(x)</math> goes to either <math>L</math> or <math>M</math>, where <math>M>L</math>. If <math>h(x)</math> decreases to <math>L</math>, then <math>f(x)</math> goes to either <math>L</math> or <math>N</math>, where <math>N<L</math>. Since <math>f(x)</math> can't go to <math>M</math> or <math>N</math>, then <math>f(x)</math> must go to <math>L</math>. Therefore, <math>\lim_{x\to S}f(x)=L</math>.
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For all <math>\varepsilon >0</math>, we must prove that there is some <math>\delta > 0</math> for which <math>|x-S|<\delta \Rightarrow |f(x)-L|<\varepsilon</math>.
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Now since, <math>\lim_{x\to S}g(x)=\lim_{x\to S}h(x)=L</math>, there must exist <math>\delta_1,\delta_2>0</math> such that,
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 +
<math>|x-S|<\delta_1 \Rightarrow |g(x)-L|<\varepsilon</math> and,
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<math>|x-S|<\delta_2 \Rightarrow |h(x)-L|<\varepsilon.</math>
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Now let <math>\delta = \min\{\delta_1,\delta_2\}</math>. If <math>|x-S|<\delta</math>, then
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<math>-\varepsilon < g(x) - L \leq f(x) - L \leq h(x) - L < \varepsilon.</math>
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So <math>|f(x)-L|<\varepsilon</math>. Now by the definition of a limit, we get <math>\lim_{x\to S}f(x)=L</math>.
  
 
==See Also==
 
==See Also==

Revision as of 20:52, 4 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 $\lim_{x\to S}f(x)=L$.

See Also

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