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The Squeeze Theorem


Date: 10/12/98 at 18:57:18
From: C Shin
Subject: Calculus - Squeeze Theorem

Dear Dr. Math:

I have a question regarding the Squeeze Theorem and using this theorem 
to solve this question:

   If f(x) = x sin(1/x), what is lim x->0 f(x)?

I already know that I need g(x) <= f(x) <= h(x), but which functions 
should I consider for g(x) and h(x)? In the explanation for this 
problem, they use absolute values for each function and use 
inequalities. Why are the functions separated, and why are we making 
these functions to be absolute value?

I thank you in advance for your help and suggestions.


Date: 10/13/98 at 07:53:35
From: Doctor Jerry
Subject: Re: Calculus - Squeeze Theorem

Hi C Shin,

There are various forms of the Squeeze Theorem (sometimes called the 
Sandwich Theorem). Here's one: If f is bounded near a point a (f need 
not be defined at a) and g has limit 0 at a, then the limit of 
f(x)*g(x) is 0 as x->a.

Here, f(x) = sin(1/x) is bounded near a = 0 and g(x) = x has limit 0 
as x->0. So, by the Squeeze Theorem, x*sin(1/x)->0 as x->0.

The explanation you have mentioned sounds like a direct proof of the 
Squeeze Theorem, perhaps applied to the specific functions sin(1/x) 
and x.

To prove the squeeze theorem, let E > 0 be given. We are looking for a 
D > 0 for which -E < f(x)*g(x) < E whenever 0 < |x-a| < D.

Okay, if f is bounded near a, then there are numbers M and D1 > 0 for 
which |f(x)| < M whenever 0 < |x-a| < D1.

With E > 0 given, calculate E2 = E/M. Since lim g(x) = 0 as x->a, we 
can find D2 >0 for which -E2 < g(x) < E2 whenever 0 < |x-a| < D2.

Now choose D to be the minimum of D1 and D2. Then, if 0 < |x-a| < D:

   |f(x)*g(x)| < M*|g(x)| < M*E2 = M*(E/M) = E

This proves that f*g->0 as x->a.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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