Difference between revisions of "Squeeze Theorem"

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(There's no need for the inequalities to be strict)
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==Proof==
 
==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 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>.
 
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>.

Revision as of 20:40, 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.

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

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