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Squeeze Law (help)

  1. Oct 10, 2004 #1
    need someone to explain this law to me. i understand the fact that if a function exists between two other functions on a graph then it can be squeezed hence the "squeeze". What i dont understand is how it prooves this.
     
  2. jcsd
  3. Oct 10, 2004 #2

    Galileo

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    The squeeze theorem I`m familiar with says that if:
    [tex]\lim_{x \rightarrow a}f(x)=b[/tex]
    [tex]\lim_{x \rightarrow a}g(x)=b[/tex]
    and [itex]f(x)>h(x)>g(x)[/itex] for all x in a certain neighbourhood of a.
    Then
    [tex]\lim_{x \rightarrow a}h(x)=b[/tex]

    The proof follows easily from the definition of a limit.
    If [itex]|g(x)-g(a)|<\epsilon[/itex] and [itex]|f(x)-f(a)|<\epsilon[/itex],
    what does that mean for h(x)?
     
  4. Oct 10, 2004 #3
    ok. that makes sense. now how can you apply that to say g(x)=(xsin(1/x)) x cannot = 0.

    would abs(x) be your two functions that squeezes g(x)? then if the limits of
    -x and +x are the same then the limit g(x) must also be this?

    and one last question, how do you know when to use this law?
     
  5. Oct 10, 2004 #4
    its also callled the sandwich theorem lol
     
  6. Oct 11, 2004 #5

    Galileo

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    Not sure what you mean exactly, but I think you have the right idea.
    Anyway, don't use [itex]\pm|x|[/itex], since it doesn't bound your function.
    Suppose you want to show that [itex]lim_{x
    \rightarrow 0}x^2\sin(1/x) =0[/itex]
    Because [itex]-1\leq sin(1/x) \leq 1[/itex], we have [itex]-x^2\leq x^2sin(1/x) \leq x^2[/itex].
    Then apply the squeeze theorem.

    There are no rules for when to use it. But it's commonly used when there's an oscillating and bounded term like here.
     
    Last edited: Oct 11, 2004
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