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) | + | 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.
Theorem
Suppose is between and for all in the neighborhood of . If and approach some common limit L as approaches , then .
Proof
If is between and for all in the neighborhood of , then either or for all in the neighborhood of . Since the second case is basically the first case, we just need to prove the first case.
If increases to , then goes to either or , where . If decreases to , then goes to either or , where . Since can't go to or , then must go to . Therefore, .