AoPSWiki
Preparing for MATHCOUNTS or the AMC contests, and having a tough time with number theory problems? Read Art of Problem Solving's Introduction to Number Theory by Mathew Crawford.
Personal tools

Squeeze Theorem

From AoPSWiki

The Squeeze Theorem (also called the Sandwich Theorem or the Squeeze Play Theorem) is a relatively simple theorem that deals with calculus, specifically limits.

Error creating thumbnail: sh: /usr/bin/convert: No such file or directory
Squeeze Theorem

Contents

Theorem

Suppose f(x) is between g(x) and h(x) for all x in a neighborhood of the point 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 this neighborhood. The two cases are the same up to renaming our functions, so assume without loss of generality that g(x)\leq f(x) \leq h(x).

We must show that for all \varepsilon >0 there is some \delta > 0 for which |x-S|<\delta implies |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 \textrm{  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 as desired.

Applications and examples

This section is incomplete. You can help us out by completing it.


See Also

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us