Difference between revisions of "Squeeze Theorem"

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{{WotWAnnounce|week=May 4-11}}
 
{{WotWAnnounce|week=May 4-11}}
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.
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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.
  
 
[[Image:Squeeze theorem example.jpg|thumb|Squeeze Theorem]]
 
[[Image:Squeeze theorem example.jpg|thumb|Squeeze Theorem]]

Revision as of 19:07, 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 $g(x)<f(x)<h(x)$ or $h(x)<f(x)<g(x)$ 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.

If $g(x)$ increases to $L$, then $f(x)$ goes to either $L$ or $M$, where $M>L$. If $h(x)$ decreases to $L$, then $f(x)$ goes to either $L$ or $N$, where $N<L$. Since $f(x)$ can't go to $M$ or $N$, then $f(x)$ must go to $L$. Therefore, $\lim_{x\to S}f(x)=L$.

See Also