Difference between revisions of "Squeeze Theorem"
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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)<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>. The second case is basically the first case, so 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)<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>. The second case is basically the first case, so 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>. |
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==See Also== | ==See Also== | ||
*[[Limit]] | *[[Limit]] | ||
Revision as of 12:10, 1 December 2007
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 . The second case is basically the first case, so 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, .
