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Homework Help: Find the limit

  1. Apr 22, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    [itex]\lim_{x->0} x\sin\frac{1}{x} [/itex]


    3. The attempt at a solution

    I can rewrite the function as [itex]\dfrac{\sin\frac{1}{x}}{1/x} [/itex]

    But this can't be equal to 1 as the argument tends to infinity.
     
  2. jcsd
  3. Apr 22, 2014 #2

    LCKurtz

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    Hint: How big can ##|\sin \frac 1 x|## be?
     
  4. Apr 22, 2014 #3

    utkarshakash

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    It will oscillate between -1 and 1. So the maximum absolute value can be 1.
     
  5. Apr 22, 2014 #4

    LCKurtz

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    So....?
     
  6. Apr 22, 2014 #5
    I would use [itex] \lim_{x\rightarrow a} (f(x) \cdot g(x)) = \lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x) [/itex] to tackle this problem personally.
     
  7. Apr 22, 2014 #6

    LCKurtz

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    Bad idea.
     
  8. Apr 22, 2014 #7
    Oops! :redface: Sorry, you can't use that property. I will go hang my head in shame.
     
  9. Apr 22, 2014 #8

    utkarshakash

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    But what about 1/x contained in the denominator? It tends to infinity as x approaches zero. How do I find limit then?
     
  10. Apr 22, 2014 #9

    LCKurtz

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    Your original problem is ##x\sin\frac 1 x##. Use that.
     
  11. Apr 22, 2014 #10

    ChrisVer

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    it's better to think about:
    [itex]lim_{x\rightarrow 0^{+}} f(x) = lim_{x\rightarrow 0^{-}} f(x)=A[/itex]
    then [itex] lim_{x\rightarrow 0} f(x)=A[/itex]
     
  12. Apr 23, 2014 #11

    utkarshakash

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    When x approaches 0 sin(1/x) approaches 1. So the limit should be 0, right?
     
  13. Apr 23, 2014 #12
    The answer is correct but the reasoning is not, IMO.

    As x approaches 0+ (or 0-), 1/x approaches infinity (or -infinity) but sin(1/x) approaches some value between -1 and 1 which is a finite number and a finite number multiplied with 0 is zero.
     
  14. Apr 23, 2014 #13

    utkarshakash

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    But the correct answer is 1 according to the solution. :frown:
     
  15. Apr 23, 2014 #14
  16. Apr 23, 2014 #15

    Curious3141

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    Then that "correct answer" is incorrect. The limit of your expression is 0.

    But if the limit is taken as x →∞, then the answer is indeed 1. You sure this wasn't the question?
     
  17. Apr 23, 2014 #16

    utkarshakash

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    I too think the given answer is incorrect.
     
  18. Apr 23, 2014 #17

    LCKurtz

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    @utkarshakash: The correct answer is zero, but you haven't given a good argument for it yet. Write down some inequalities.
     
    Last edited: Apr 23, 2014
  19. Apr 23, 2014 #18
    Just take abs(xsin(1/x))=<abs (x).
     
  20. Apr 23, 2014 #19

    utkarshakash

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    I tried to apply Sandwich Theorem. Here's my attempt:

    [itex]-1 \leq \sin \frac{1}{x} \leq 1 \\
    -x \leq \sin \frac{1}{x} \leq x \\
    lim_{x \to 0} -x = lim_{x \to 0} x = 0 [/itex]

    The limit should therefore be 0.

    Is my reasoning correct?
     
  21. Apr 23, 2014 #20

    LCKurtz

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    That only follows from the first inequality if ##x\ge 0##.

    So that would only work for ##x\to 0^+##.

    It is possible to fix that argument, but much easier to work with ##\left| x\sin(\frac 1 x)\right|##.

    [Edit] Note lurflurf's following comment about the missing ##x## which I just added.
     
    Last edited: Apr 23, 2014
  22. Apr 23, 2014 #21

    lurflurf

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    That is good except for a typo, it should read
    $$-1 \leq \sin \frac{1}{x} \leq 1 \\
    -x \leq x\, \sin \frac{1}{x} \leq x \\
    lim_{x \to 0} -x = lim_{x \to 0} x = 0$$

    edit: The inequality changes sign at x=0
     
    Last edited: Apr 23, 2014
  23. Apr 24, 2014 #22

    utkarshakash

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    [itex]0 \leq \left|\sin(\frac 1 x)\right| \leq 1 \\
    0 \leq \left|x\sin(\frac 1 x)\right| \leq |x| [/itex]

    Applying Sandwich Theorem now would give the answer 0. I think I'm correct this time .
     
  24. Apr 24, 2014 #23
    In general, whene you have a clogged function (like [itex]\sin[/itex] or [itex]\cos[/itex]) times 0 the [itex]\lim[/itex] will be 0. It is usefull to think like this! Taking the inequalities and sandwich theorem will allways help!

    I usually face these limits like this: [itex] |x \sin(\frac 1 x) | \leq |x| ⇔ -|x| \leq x \sin(\frac 1 x) \leq |x| [/itex]
    Then you apply the sandwich theorem exactly as you did. Allthough I best like your method this is supposed to work a lot more times I think. Plus you take the [itex]\lim_{x \to α} f(x)[/itex] instead of [itex]\lim_{x \to α} |f(x)|[/itex]
     
    Last edited: Apr 24, 2014
  25. Apr 24, 2014 #24

    utkarshakash

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    Can you please format your equations correctly? I can't understand it.
     
  26. Apr 24, 2014 #25
    I just edited it!
     
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