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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum