Difference between revisions of "Squeeze Theorem"

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The '''Squeeze Theorem''' (also called the '''Sandwich Theorem''' or the '''Squeeze Play Theorem''') is a relatively simple [[theorem]] that deals with [[calculus]], specifically [[limit]]s.
The '''Squeeze Play Theorem''' (also called the '''Squeeze Theorem''' or the '''Sandwich Theorem''') is a relatively simple [[theorem]] that deals with [[calculus]], specifically [[limit]]s.
 
  
 
[[Image:Squeeze theorem example.jpg|thumb|Squeeze Theorem]]
 
[[Image:Squeeze theorem example.jpg|thumb|Squeeze Theorem]]
  
 
==Theorem==
 
==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>.
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Suppose <math>f(x)</math> is between <math>g(x)</math> and <math>h(x)</math> for all <math>x</math> in a [[neighborhood]] of the point <math>S</math>. If <math>g</math> and <math>h</math> approach some common limit <math>L</math> as <math>x</math> approaches <math>S</math>, then <math>\lim_{x\to S}f(x)=L</math>.
  
==Proof==
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===Proof===
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)<f(x)<h(x)</math> or <math>h(x)<f(x)<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.
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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 this neighborhood.  The two cases are the same up to renaming our [[function]]s, so assume without loss of generality that <math>g(x)\leq f(x) \leq h(x)</math>.
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We must show that for all <math>\varepsilon >0</math> there is some <math>\delta > 0</math> for which <math>|x-S|<\delta</math> implies <math>|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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<cmath>|x-S|<\delta_1 \Rightarrow |g(x)-L|<\varepsilon \textrm{  and  } |x-S|<\delta_2 \Rightarrow |h(x)-L|<\varepsilon.</cmath>
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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> as desired.
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== Applications and examples==
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{{incomplete|section}}
  
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>.
 
  
 
==See Also==
 
==See Also==

Revision as of 21:50, 19 May 2008

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 a neighborhood of the point $S$. 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 this neighborhood. The two cases are the same up to renaming our functions, so assume without loss of generality that $g(x)\leq f(x) \leq h(x)$.

We must show that for all $\varepsilon >0$ there is some $\delta > 0$ for which $|x-S|<\delta$ implies $|f(x)-L|<\varepsilon$.

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

\[|x-S|<\delta_1 \Rightarrow |g(x)-L|<\varepsilon \textrm{  and  } |x-S|<\delta_2 \Rightarrow |h(x)-L|<\varepsilon.\]

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 $\lim_{x\to S}f(x)=L$ as desired.

Applications and examples

Template:Incomplete


See Also

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