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Limit of sin(1/x)

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate [tex]\mathop {\lim }\limits_{x \to 0 } x^2 \sin \frac{1}{x}[/tex]


    2. Relevant equations

    "product law" of limits I think


    3. The attempt at a solution

    I took a guess at 0, as 1/x would become very large as x goes to 0, so sin would become a maximum of 1 but [tex]x^2[/tex] would become 0.

    Is this actually the correct way to solve this limit problem? How should I actually express the answer, just put 0 or is there some intermediate working that can be shown?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 15, 2009 #2

    dx

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    Use the sandwich rule.

    -x2 ≤ x2 sin(1/x) ≤ x2
     
  4. May 15, 2009 #3

    tiny-tim

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    Hi username12345! :smile:

    Yes, the product law is the way …

    just say it's lim{AB}, and |B| ≤ 1 … :wink:
     
  5. May 15, 2009 #4

    matt grime

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    I think you're tex'ing or mathml'ing is off, dx. Did you mean to write

    [tex] -x^2 \le x^2\sin(1/x) \le x^2[/tex]
     
  6. May 15, 2009 #5

    dx

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    Yes, thanks matt. I corrected it.
     
  7. May 15, 2009 #6
    The text I have has a "squeeze law" which I think is the same. Looks the same.
     
  8. May 15, 2009 #7

    dx

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  9. May 15, 2009 #8
    OK I am trying to use this sandwhich rule, but it doesn't make sense to me.

    [tex]\mathop {\lim }\limits_{x \to 1 } |x-1|\sin \frac{1}{x-1}[/tex]

    By the product rule |x-1| we'd get 0 so the limit would be zero, but for fun I try squeeze.

    [tex]-|x-1| \leq \sin \frac{1}{x-1} \leq |x-1|[/tex]
    [tex]0 \leq \sin \frac{1}{x-1} \leq 0[/tex]

    so, [tex]\mathop {\lim }\limits_{x \to 1 } |x-1|\sin \frac{1}{x-1} = 0[/tex]

    What is odd to me is that as x tends to 1 [tex]\sin \frac{1}{x-1}[/tex] is not 0. For example when x is .999999999999 the result is 0.98.
     
  10. May 15, 2009 #9

    dx

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    Is this a different question?

    For the original question, technically you cannot use the product rule because

    [tex] \lim_{x \rightarrow 0} \sin \frac{1}{x} [/tex]

    doesn't exist.
     
  11. May 15, 2009 #10
    Yes, sorry, it is a different question.

    You suggested to use sandwhich so rather than make a new thread I tried to apply it to a similar problem.
     
  12. May 15, 2009 #11

    dx

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    Ok,

    Do you understand why

    -|x-1|≤ |x-1|sin(1/|x-1|) ≤ |x-1|?

    You forgot the |x-1| in the middle multiplying sin(1/|x-1|).
     
  13. May 15, 2009 #12
    With |x-1| in the middle it makes more sense. My understanding though is limited to simply substituting x = 1 and evaluating the expression. I couldn't really explain it.

    Thanks dx and to tiny-tim aswell.
     
  14. May 15, 2009 #13
    I would just say that the limit is zero because x^2 goes to zero and the function sin(1/x) is bounded.

    And I don't know if you're interested, but you can show that the limit of sin(1/x) as x->0 doesn't exist by letting f(x)=sin(1/x) and finding a sequence (Sn) that goes to zero but f(Sn) doesn't converge.
     
  15. May 15, 2009 #14

    rock.freak667

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    Shouldn't doing this work the same

    [tex]x^2sin(\frac{1}{x})=x\frac{sin(\frac{1}{x})}{\frac{1}{x}}[/tex]

    and then just use the fact that

    [tex]\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) \times \lim_{x \rightarrow a} g(x)[/tex]
     
  16. May 15, 2009 #15
    That's very clever.
     
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