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Homework Help: Using Squeeze Theorem To Solve An Obvious Problem

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    So we have to evaluate this using the squeeze theorem:
    http://www4c.wolframalpha.com/Calculate/MSP/MSP230197da409360513h500003f12d5h334he7b32?MSPStoreType=image/gif&s=56 [Broken]

    3. The attempt at a solution
    Well it's pretty obvious that it will go to zero. Reason being that anything divided by infinity is zero...but how do you evaluate this with the squeeze theorem? There's only one equation given, and it's easily solved.

    Thanks in advance.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 29, 2009 #2


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    That is a pretty lousy (or extremely sloppy formulated) reason :tongue:
    You seem to think that [tex]\lim_{x\to \infty}\frac{f(x)}{x}=0[/tex] no matter what f is. That is certainly not true (take e.g. f(x)=x, or f(x)=x^2).

    The key observation here is that the sine is bounded, in particular [tex]|\sin x|\leq 1[/tex] for all [tex]x[/tex], in other words [tex]-1\leq \sin x\leq 1[/tex]. Try to use this.
  4. Sep 29, 2009 #3
    Hmm, the image of the function vanished.

    So you're saying I should use -1 and 1 as my other two functions that go to infinity to evaluate sine at infinity?
  5. Sep 29, 2009 #4
    Since sinx only goes between 1 and -1, it will never make sinx/x go higher or lower than 1/x or -1/x, respectively. Can you understand why?
  6. Sep 29, 2009 #5
    I guess. Because trig functions have a range equal to that of +-their amplitude?
  7. Sep 29, 2009 #6
    The amplitude of sinx is 1. Will numbers between 0 and 1 (ignoring the negative values of sinx) multiplied with 1/x ever make 1/x be greater than it normally is?
    After than, consider the negative values of sinx and -1/x.
  8. Sep 30, 2009 #7


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    I don't quite understand what you mean by this.
    No, you don't need to 'evaluate sine at infinity' (this not a meaningful expression).

    Could you please state the squeezing theorem? Then you will immediately see what to do with the estimate of sine being between -1 and 1.
  9. Sep 30, 2009 #8
    I just got it, thanks for the idea bouncing guys. It should be evaluated as "-1/x <= sin(x) <= 1/x".

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