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Limits: Evaluating function of 2-variables where a limit doesn't exist

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate:

    [tex]\lim_{(x,y)\to(0,0)} \sqrt{x^2+y^2}\sin \frac{1}{\tan xy}[/tex]

    **Apologies that this Tex didn't come out, I can't see where the typos are**Hopefully you can still determine the function I am trying to write**





    3. The attempt at a solution

    So, I can see this isn't solveable by just plugging in the co-ords as the sine function will not be valid. I can't separate them as the Sine function isn't continous due to Tan (xy) being in the denominator. I can't fix x or y or substitute a function for y = g(x).

    I can't see a way forward or another example that would give me the direction I'm lacking.

    Any pointers guys?
     
    Last edited by a moderator: Nov 8, 2012
  2. jcsd
  3. Nov 8, 2012 #2

    Zondrina

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    I believe that your function is :

    [itex]lim_{(x,y)\to(0,0)} \sqrt{x^2+y^2} sin(\frac{1}{\tan(xy)})[/itex]
     
  4. Nov 8, 2012 #3
    Yep. Noted the differences. Thanks
     
  5. Nov 8, 2012 #4

    Zondrina

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    What do you know about the squeeze theorem? What do you know about |sinx| ?
     
  6. Nov 8, 2012 #5
    Not heard of it before. Just watched a tutorial online and an interesting theorem. It was only in one variable and don't see much out ther on 2-variable functions. Not sure about the modulus of sin(x)? It's alawys positive?
     
  7. Nov 8, 2012 #6

    SammyS

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    The function, [itex]\displaystyle f(x,\,y)= \sqrt{x^2+y^2}\sin(\frac{1}{\tan(xy)})[/itex] is a mess near (0, 0).


    Perhaps the problem is: find [itex]\displaystyle \lim_{(x,y)\to(0,0)}\sqrt{x^2+y^2}\sin(\tan^{-1}(xy))\ ?[/itex]
     
  8. Nov 8, 2012 #7
    No the original equation is correct. Definately not arctan!
     
  9. Nov 8, 2012 #8

    Zondrina

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    Like I said, what do you know about |sin(x)| for any x.
     
  10. Nov 8, 2012 #9

    SammyS

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    OK.

    Putting arctan in there would make the problem uninteresting .

    Zondrina has the right approach anyway .
     
  11. Nov 8, 2012 #10
    The only thing i can say about it is its always positive... not sure how that helps as it's not the modulus in the equation and1/ tan(xy) will alwys be undefined at (0,0)???
     
  12. Nov 8, 2012 #11

    Zondrina

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    That's not it. Think about the graph of sin(x). How does it relate to |sin(x)|?
     
  13. Nov 8, 2012 #12

    SammyS

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    What is the range of sin(x) ?
     
  14. Nov 8, 2012 #13
    the range for sin(x) is [-1,1] and for the modulus is [0,1].

    Could you be a bit more explicit with where to go with this. How do you deal with the tangent in the denominator with this?
     
  15. Nov 8, 2012 #14

    Zondrina

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    Exactly, the range of sin(x) is the interval [-1,1]. Now what does this tell you about |sin(x)|??
     
  16. Nov 8, 2012 #15
    I'm sorry I just don't know other that its interval is [0,1]...?
     
  17. Nov 8, 2012 #16

    Zondrina

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    Do you understand that |x| < 1 implies that -1 < x < 1?

    Apply this concept to |sin(x)|
     
  18. Nov 8, 2012 #17
    Yes, I understand the inequalities you mention but I have not been taught how to manipulate the original question using this method.

    Thanks for your help though. I'll try to catch one of my tutors or classmates to help explain.

    P.
     
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