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How to take the limit of a multivariate function that goes to 0

  1. Jun 5, 2009 #1
    Title says it all.

    Examples from class I am looking at include:

    lim(x,y) -> (0,0) of functions such as:

    6yx^3 / 2x^4 + y^4

    or

    (x^2)(sin^2 (y)) / x^2 + 2y^2

    My professor did a bad job of explaining it (or at least I did not understand).

    Thanks,
    Nkk
     
  2. jcsd
  3. Jun 5, 2009 #2
    You'll need to try evaluating the limit along different curves, ie. try setting y=0 and evaluating, try setting x=0 and evaluating, try setting y=x and evaluating, (etc depending on the problem). If the limits along any of the curves that you try are different, then the limit does not exist. The function only has a limit only when the values of f(x,y) approach the same number no matter how (x,y) approaches (a,b) (in your case a,b = 0,0).
     
  4. Jun 5, 2009 #3

    HallsofIvy

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    If (x,y) approaches (0, 0) along the x-axis- that is, as (x, 0), the two functions above become 6(0)(x^3)/(2x^4+ 0)= 0 and x^2(sin^2(0))/(x^2+ 2y^2)= 0 for all x so the limit as (x, 0)-> (0, 0) is 0. But in order that the limit itself exist, we must get the same result no matter which path we use to approach (0, 0). On the line y= x the first function becomes 6x(x^3)/(2x^4+ x^4)= x^4(5x^4)= 1/5 so no matter how close to (0,0) we get on that line, the value of the function is 1/5 and does not approach 0.

    More generally, the best thing to do is to convert to polar coordinates. That way, a single variable, r, measures the distance from (0,0) no matter what the other, [itex]\theta[/itex], is. (In Cartesian coordinates distance is a combination of x and y.)
    [tex]\frac{6yx^3}{2x^4 + y^4}= \frac{r sin(\theta)r^3cos^3(\theta)}{2r^4cos^4(\theta)+ r^4 sin^4(\theta)}[/tex]
    [tex]= \frac{sin(\theta)cos^3(\theta)}{2cos^4(\theta)+ sin^4(\theta)}[/tex]
    Notice that the "r" terms have cancelled out- the value, no matter how close to (0,0) we are, depends upon [itex]\theta[/itex]. Again, there is no one value we approach as r goes to 0 and so no limit.

    For the second function,
    [tex]\frac{x^2sin(y)}{x^2+ 2y^2}= \frac{r^2cos^2(\theta)sin(rcos(\theta))}{r^2 (cos^2(\theta)+ 2sin^2(\theta)}[/tex]
    [tex]= \frac{cos(\theta)sin(rcos(\theta)}{cos^2(\theta)+ 2sin^2(\theta)}[/tex]
    Now, as r goes to 0, what happens to [itex]rcos(\theta)[/itex] and then [itex]sin(r cos(\theta))[/itex]?
     
  5. Jun 7, 2009 #4
    SO on the last example, since rcos(theta) goes to 0 as r goes to 0, and this you get sin(0), the limist is 0. That is clear with the polar coordinates.

    However, for the first example, you say that there is not an angle that theta goes to as r goes to zero, but why? I understand that at r=0 there is no defined theta, but as r goes to zero (example at 10^-100000), with r pointing from the origin to the curve, there is a theta, so would not theta approach the ange that is right after r=0? Does that make any sense? I know what you are saying is correct as it intuitively makes sense, but I cannot reason through it, which is causing problems.

    I know I am being dense about something, so thanks for taking the time to explain this.

    -Nkk

    EDIT: also, if you convert x,y to r, theta, why would you not count the theta and take it to some number such as arctan (y/x), as per polar coordinate conversions? That would bring the first example to 0/2, which indicates it exists to be zero, not that it does not exist. What is wring with my logic of making theta into arctan (y/x)? I think understanding why we can discount theta is my problem.
     
  6. Jun 7, 2009 #5

    HallsofIvy

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    If [itex]\lim_{r\to 0} f(r, \theta)[/itex] depends on [itex]\theta[/itex], because there are points arbitarily close to 0 with r close to (0,0) but different values of [itex]\theta[/itex], we would have points arbitrarily close to (0,0) giving different values of the function. That contradicts the definition of "limit".

    Because you can't choose a particular value of [itex]theta[/b] that holds for all points close to 0! Yes, if you approach (0,0) along the line y= 0 ([itex]\theta[/itex]= 0) you get 0: there exist points arbitrarily close to 0 such that f(x,y) is close to 0. Similarly, if you approach (0,0) along the line x= 0 ([itex]\theta= \pi/2[/itex]) you get a "limit" of 0.
    But if you approach (0,0) along the line y= x ([itex]\theta= \pi/4[/itex]), you get a "limit" of 2 which means there are points arbitrarily close to (0,0) for which the function is close to 2. There is no "neighborhood" (small disk) around (0,0) for which all points in that disk give values of the function close to a specific number: there is no limit.
     
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