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L'Hôpítal's rule for multivariate functions!

  1. Nov 17, 2008 #1
    In my math class lectures at the university while studying multivariable functions the lecturer never mentioned L'hopital's rule for these multivariate functions..But in a tutorial class,a tutorial assistant approached this question..find lim (x,y)-->(0,0) [sin(x^2+y^2)]/(x^2+y^2)..by implicitly differentiating the numerator and denominator to get [xcos(x^2+y^2)+ycos(x^2+y^2)]/(x+y)...{mind you,I know,there are no indications for dx and dy,thats the approach I am just quoting it down!] then there was more implicit differentiation [the aim was as usual to remove variability of the denominator function]..the result was [cos(x^2+y^2)-2x^2sin(x^2+y^2)-4xysin(x^2+y^2)-2y^2sin(x^2+y^2)+sin(x^2+y^2)]/2...again no dx,dy or whateva!but by approach a limit lim (x,y)-->(0,0) you get 1...which is obviously the actual answer!I know this so because anyways I can the function f(x,y)=z and change the limit from lim (x,y)->(0,0) to lim z->0 so the whole thing is lim z->0 (sin z)/z which is a mathematical fact to be equal to 1 and can prove so by l'hopital's rule for single variable functions,sandwich theorem and by taylor's theorem!...am not here to disprove my tutor,i just want to know the l'hopital's rule for multivariate function,for personal pleasure,so i dont necessarily need completely proven facts and textbook quotes,I need a real discussion!I welcome eccentric thoughts and suggestions...i've been searching through google and got dropped here less than a week ago...Here's my thought,no proof,just speculation and am still looking into the facts to perfect my ideas,For a single variable function,l'hopital's rule differentiates a function f(x) with respect to x,so am thinking with multivariable functions I would take partial derivatives of first order followed by the mixed derivative...this in my case for now will be abstractly to show if a case is determinate or indeterminate,not that the result would be the real value itself,though it might be but not necessarily!You might not understand this so please dont take the trouble to post negative messages,am expecting an academically progressive discussion thank you!
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  3. Nov 17, 2008 #2


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    One way of understanding why L'hôpital's rule works is through differential approximation -- you can write any differentiable function as:
    [tex]f(x) = f(0) + xf'(0) + xe(x)[/tex]
    where e has the property that
    [tex]\lim_{x \rightarrow 0} e(x)[/tex] = 0
    (This same fact can be derived via Taylor series, and a similar one via the mean value theorem)

    Try deriving (a form of) L'hôpital's rule from this.

    Now, how does this work in two dimensions?
  4. Nov 17, 2008 #3
    I recognise the approximation function...Still dont quite get it,but I'll work on...Am still pondering on how l'hopital came up with that rule in the 1st place,if I would know the roots of this rule I'd be to extrapolate to other conditions...This is not a homework question,if you have the formula,its derivation or proof I'd appreciate if you post it over.
  5. Nov 19, 2008 #4


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    Did you try plugging the differential approximation in the limit of [itex]f(x)/g(x)[/itex]?

    The same principle is at play, though (just without the cheating issue compounded on top of it) -- you'll understand it far better if you work it out yourself than if you see someone else do it.

    Historically, L'Hôpital didn't invent his rule; he was given permission to attach his name to it, or something like that. I don't know how it was originally discovered, but the spirit of calculus is the art of approximating things. Things like differential approximations, mean value theorem, Taylor series... these are all basic tools of approximation. So if you know the value of f(0), and f is differentiable there, it is very natural to try and use your knowledge of f(0) to construct an approximation of f(x) that is good near 0.

    Incidentally, one form of 2-variable differential approximation is:

    [tex]f(x, y) = f(0) + x f_x(0, 0) + y f_y(0, 0) + e(x, y)[/tex]
    [tex]\lim_{(x, y) \to (0, 0)} \frac{e(x, y)}{||(x, y)||} = 0[/tex]

    (I hope I have that right) It looks prettier as

    [tex]f(\vec{v}) = f(\vec{0}) + \nabla_{\vec{v}}f(0, 0) + e(\vec{v})[/tex]
    [tex]\lim_{\vec{v} \to \vec{0}} \frac{e(\vec{v})}{||\vec{v}||} = 0[/tex]
    Last edited: Nov 19, 2008
  6. Nov 19, 2008 #5
    I dont cheat,I've never cheated,this not an exam,am never going to be asked to derive this!But thanks for the insight...Yes am checking it through,seems to go in the right direction but it wont get me there...The reason why I need the historical background of this formula is because I'd like to know why?Why they came up with this rule?Knowing the roots allows me better knowledge,am still looking through!Thanks for your post,really helpful!
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