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Squeeze Theorem

  1. Sep 18, 2007 #1
    Hey could someone explain the squeeze theorem to me a little; I understand you want 2 equations that "squeeze" another one into between them sothat you can find they're limits and find the equations limit but how do you find the 2 equations that squeeze the original one in?
  2. jcsd
  3. Sep 19, 2007 #2
    There's no general way of finding two functions that satisfy the criteria...
    You just want to find two functions [tex]g[/tex] and [tex]h[/tex] such that:
    [tex]lim_{x\rightarrow a}g(x)=lim_{x\rightarrow a}h(x)[/tex] and
    [tex]g(x)\leq f(x) \leq h(x)[/tex] for all [tex]x[/tex] within ssome neighbourhood of [tex]a[/tex]. Then, the squeeze theorem tells you that the limit those two have at [tex]a[/tex] is the same as the limit [tex]f[/tex] has at [tex]a[/tex].
    Like I said-- the choice of your [tex]g[/tex] and [tex]h[/tex] is completely arbitrary-- you just want them to satisfy those conditions.
    Sometimes, e.g. when you have [tex]\frac{1}{x}sin(x)[/tex], boundedness helps. Observe:
    [tex]-1 \leq sin(x) \leq 1[/tex] (Property of the sine function)
    This implies [tex]-\frac{1}{x} \leq \frac{1}{x} sin(x) \leq \frac{1}{x}[/tex]. (Whenever [tex]x>0[/tex])
    As you can see, the limit as [tex]x\rightarrow \infty[/tex] of the left and right hand sides of the inequality match and equal 0, and so the limit of the function [tex]\frac{1}{x} sin(x)[/tex] is 0. This is a classic application of the squeeze theorem.
    I hope that helped.
  4. Sep 26, 2007 #3
    I get that part of it, but my question is on the 2nd part of the squeeze theorem.

    It states abs(g(x))<=M for all x not zero.
    then lim f(x) x g(x)=0 as X--> a. Can someone show me the proof to this part and explain it. The book shows no further information and I'm confused as to what it means.
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