1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Multivariable Calculus: Limits

  1. Sep 16, 2007 #1
    1) lim [x(y^2)] / (x^2 + y^2)
    (x,y)->(0,0)
    Find the value of the given limit, if it exists.

    Using polar coordinates, set x = r cos(theta), y = r sin(theta)
    Then, the given limit = lim [r cos(theta) r^2 sin^2(theta)] / r^2
    r->0
    = lim r [cos(theta) sin^(theta)]
    r->0
    = 0 since cos(theta) sin^(theta)<=1, i.e. bounded

    If I found that the limit is equal to 0 this way, can I conclude immediately that the original limit is 0 too?

    What I believe is that for the part using polar coordinates r->0, it seems that it's appraoching the origin through straight lines paths ONLY, however my textbook says that "for the limit to exist, we must get the same result no matter which of the infinite number of paths is chosen"


    Thanks for answering!
     
  2. jcsd
  3. Sep 16, 2007 #2
    If I found that the limit is equal to 0 this way, can I conclude immediately that the original limit is 0 too?

    yes you can


    i believe when you switch to polar


    and after simplification

    plug in your r

    if your equation is not a function of theta, then that is the limit


    if your equation is still a function of theta, then the limit DNE



    and you can also use lhopitals rule once you switch to polar
    if the conditions are met
     
    Last edited: Sep 16, 2007
  4. Sep 16, 2007 #3

    HallsofIvy

    User Avatar
    Science Advisor

    No, you are not approaching "through straight lines only" because you are not assuming that [itex]\theta[/itex] is a constant. You need to show that the limit goes to 0 as r goes to 0, no matter what [itex]\theta[/itex] is. The real point is that, in polar coordinates, r measures exactly the distance to the origin, irrespective of [itex]\theta[/itex]- and that is what must be made "less than [itex]\delta[/itex]".
     
  5. Sep 16, 2007 #4
    "You need to show that the limit goes to 0 as r goes to 0, no matter what theta is" <---but this covers only the paths of all the straight lines through the origin, how about in the paths of parabolas and cube root function through the origin, etc?
     
  6. Sep 16, 2007 #5
    Because theta can be a function of r...
     
  7. Sep 16, 2007 #6
    But how does this help? Sorry, I am not getting it...
     
  8. Sep 16, 2007 #7
    It's a straight line iff theta is a constant. But the proof allows theta to vary arbitrarily, therefore, you're allowed curves.
     
  9. Sep 16, 2007 #8
    How can it be a curve if theta can vary?
     
  10. Sep 16, 2007 #9
    Consider the equation [tex]\theta = r[/tex]. Plot it. Is it not a curve?
     
  11. Sep 16, 2007 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    It seems you are unsure of what theta really is.

    So I have to ask:

    What, precisely, does the variable theta measure?
     
  12. Sep 16, 2007 #11
    theta is a counterclockwise measure of "angle" from the positive x-axis, this is what makes me think of r->0 as (x,y) approaching (0,0) through straight lines only
     
  13. Sep 16, 2007 #12
    Did you try and plot the function \theta = r as I suggested? It is a straight line?
     
  14. Sep 16, 2007 #13
    Do you mean [r cos(r) r^2 sin^2(r)] / r^2 ?

    By the way, we are taking the limit r->0, shouldn't theta be fixed?
     
  15. Sep 17, 2007 #14

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Why?
    Any particular POINT can be uniquely specified by its distance from the origin, and the angle the line segment between the origin and the point makes with the positive x-axis.

    By no means does this entail that all the points a particular curve consists of make the same angle to the positive x-axis with their respective line segments.
     
  16. Sep 17, 2007 #15
    No! That's the point! We DON'T have to fix theta!
     
  17. Sep 17, 2007 #16

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You can approach the origin by way of a straight line, but also by spiralling yourself inwards. And in many other ways as well.
    Only if you approach the originalong a straight line will theta be constant.
     
  18. Sep 18, 2007 #17
    If theta is allowed to vary, will r->0 using polar coordinates cover ALL paths for which (x,y)->(0,0) ?
     
  19. Sep 18, 2007 #18
  20. Sep 18, 2007 #19

    HallsofIvy

    User Avatar
    Science Advisor

    "Paths", in polar coordinates, depend only on r and [itex]\theta[/itex]. If r ranges from, say, 1 to 0, while [itex]\theta[/itex] is allowed to have any value, then, yes, it covers all paths. More to the point is that in polar coordinates, r alone measures the distance from the origin which is what you want: ||(x,y)- (0,0)||= r< [itex]\delta[/itex].
    As long as the limit, as r goes to 0, is a number, that is, independent of [itex]\theta[/itex] that number will be the limit.
     
    Last edited by a moderator: Sep 20, 2007
  21. Sep 18, 2007 #20
    I think part of the confusion might be this: a path near the origin, if it goes through it, can be well approximated by a straight line. Unfortunately, whilst that's intuitively true, analysis can always come back to bite us:

    [tex]\theta = \pi sin(1/r)[/tex]
     
  22. Sep 20, 2007 #21
    1) Using polar coordinates, how can it cover the paths of say, parabolas, in approaching (0,0) ? (if theta can vary)

    Say, if I am trying to eavluate a limit by changing it to polar coordinates, and I get a finite limit L using the polar coordinates, can I ALWAYS immediately conclude that the original limit is L in any case like this?
     
  23. Sep 20, 2007 #22
    Let's just get things clear. The definition of limits require a metric space. In R^n, there's the obvious euclidean metric. That means when you evaluate it, you convert to a "polar" representation (or more accurately, hyperspherical). However, that's not the only one possible -- you've got the infinite family of L^n norms, and that's just in R^n.

    So, in R^2, in the case that you're using the euclidean norm (as you are), the definition of limit requires that you convert to polar coordinates.

    The question of finding the polar representation of y=x^2 is left as an exercise.
     
  24. Sep 20, 2007 #23

    HallsofIvy

    User Avatar
    Science Advisor

    If, say, the parabola is y= ax2, replacing y by [itex]r sin(\theta)[/itex] and x by [itex]r sin(\theta)[/itex], then the parabola becomes [itex]r sin(\theta)= r^2 cos^2(\theta)[/itex] or [itex]r= tan(\theta)sec(\theta)[/itex]. Now, with that relationship between r and [itex]\theta[/itex], as r goes to infinity, the point (x,y) will go to (0,0) along that curve. But, of course, the whole point is to allow r itself to go to 0, without regard for what [itex]\theta[/itex] is.

    Yes, IF that finite limit is as r goes to 0 and does NOT depend on [itex]\theta[/itex].

    For example, if f is any function, [itex]r f(\theta)+ 5[/itex], as r goes to 0, goes to 5 whatever [itex]\theta[/itex] is- if you had a function that reduced to that in polar coordinates, its limit at (0,0) would be "5". But [itex]r+ 5f(\theta)[/itex], as r goes to 0, goes to [itex]5f(\theta)[/itex] and so the value very close to (0,0) depends upon [itex]\theta[/itex] and so a function that reduced to that in polar coordinates would NOT have a limit at (0,0).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook