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Homework Help: Multi variable limits

  1. Jun 10, 2008 #1
    Evaluate the limit or determine that it DNE

    lim as (x,y) goes to (0,0) of [(sinx)(siny)]/(xy)

    3. The attempt at a solution

    I can solve these when the denominator isn't zero, but pretty much rest of my homework deals with that situation. I'm really not sure how you solve multi-variable limits when you get a zero in the denominator. I know that with single variable limits you use L'hospital, but I'm not sure if that is possible in these situations. So I really need help with this.
  2. jcsd
  3. Jun 10, 2008 #2
    let x=y, then lim as x,x -> 0,0
  4. Jun 10, 2008 #3


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    Try to get rid of either one of variables by using a suitable substitution. For example, you need not approach the origin along the y or x-axis. Try approaching the origin along the line y=mx. Make that substitution and let x approach 0, then find that limit. Try evaluating the limit in polar coordinates as well, and r approach 0 regardless of the value of theta. What do you get for both?
  5. Jun 10, 2008 #4


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    A multivariable limit exists at a point if you find the same value by approaching that point from any path. (It's analogous to, but more complicated than, the situation with one variable, where you simply have to have the same limit as you approach the point from either side.

    Do you know the limit of (sin x)/x as x approaches 0? This function could be written as a product of two such functions. Would the limit give the same value no matter how you chose to approach x = 0 and y = 0? What would that limit be?
  6. Jun 11, 2008 #5


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    Letting y= x and then x-> 0, or more generally, letting y= some function of x and then x-> 0, does NOT prove that limit exists. It is quite possible for the limit to exist along one curve through (0,0) and then not exist along another, or exist but have a different limit. Of course, if you could take the limit along two different paths and either show that the limit along one does not exist or that the two paths give different limits, they you would have shown that the limit does not exist.

    Unfortunately, if you took the limit along many different paths and always got the same limit, that would not prove that some path you hadn't tried wouldn't give a different limit.

    Generally, with limits going to (0,0) where you suspect the limit does exist (maybe you tried several paths and always got the same thing) the best thing to do is to change to polar coordinates. That way, the "nearness to (0,0)" is completely determined by r. If the limit as r goes to 0 is independent of [itex]\theta[/itex], then the limit exists and that value is the limit.

    HOWEVER, in a very simple case like this you can simply "separate" the variables:
    [tex]\lim_{(x,y)\rightarrow (0,0)}\frac{sin(x)sin(y)}{xy}= \left[\lim_{x\rightarrow 0}\frac{sin(x)}{x}\right]\left[\lim_{y\rightarrow 0}\frac{sin(y)}{y}\right][/tex]
    And surely you know those two limits from Calculus I!

    (You will soon be seeing much more difficult limits than this!)
  7. Jun 11, 2008 #6


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    Under what circumstances would it be valid to separate the mutivariable limits and evaluate them individually?
  8. Jun 11, 2008 #7
    - I would also like to know when you're able to use the product rule for these types of problems. I think my biggest problem right now is that I'm not sure what techniques are legal for these problems. Calculus has been pretty simple for me until this class, so it's frustrating that I haven't been able to just teach myself rather quickly.
  9. Jun 11, 2008 #8
    Product rule? Ivy simply splitted the terms in a x-part and a y-part.

    Try this one:

    [tex] \lim_{ (x,y) \rightarrow (0,0)}\ \frac{x+y}{ \sqrt{x^2 + y^2}} [/tex]
  10. Jun 11, 2008 #9
    - Sorry, read that part wrong in the book. I meant to ask when you could treat it as a product of functions of one variable. I'm about to head to class, so I'll try to work on that problem while I'm waiting for class to start.
  11. Jun 11, 2008 #10
    What Ivy did is generally speaking not always possible. If you can split the limit in one part only dependent on x and another one dependent on y you can try to take the limit of those terms seperately.
  12. Jan 15, 2010 #11
    Hey when you use polar coordinates here. What about the value of theta, my sir was telling something about the value of theta between 0 and 2 pi
  13. Aug 3, 2010 #12
    All of the normal limit laws happen with multivariables? Would that be usefull?
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