Squeeze Theorem

Revision as of 20:52, 4 May 2008 by Jam (talk | contribs) (The old proof implicitly assumed that f converged. Replaced this with an epsilon-delta proof.)
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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 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.

For all $\varepsilon >0$, we must prove that there is some $\delta > 0$ for which $|x-S|<\delta \Rightarrow |f(x)-L|<\varepsilon$.

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,

$|x-S|<\delta_1 \Rightarrow |g(x)-L|<\varepsilon$ 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 $\lim_{x\to S}f(x)=L$.

See Also

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