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0/0 as the result of taking a limit

  1. Dec 17, 2004 #1

    quasar987

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    I know when we encounter 0/0 as the result of taking a limit, we cannot conclude to the value of the limit. But what about 0/0 for a function? I am confused because in last year's final calculus exam, there's a question that reads:

    Calculate the gradient vector of the function

    [tex]f(x,y)=\frac{x^4+y^3}{x^2+y^2}[/tex]

    at all point (including at (0,0)).

    But for exemple,

    [tex]\left[ \frac{\partial f}{\partial x} \right]_{(0,0)} = \left[\frac{4x^3}{x^2+y^2}-\frac{2x^5}{(x^2+y^2)^2}-\frac{2xy^3}{(x^2+y^2)^2}\right]_{(0,0)} = \frac{0}{0}[/tex]
     
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  3. Dec 17, 2004 #2

    dextercioby

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    You missunderstood a great deal of mathematics.When u're given a function,u have to see first at which points (what set) it is defined.In your case,neither the function,nor the derivatives are defined in (0,0),so your calculation is pointless.I should say INCORRECT.

    Daniel.

    PS.Was it written on the exam paper "including (0,0)"??If so,it must have been a (bad) joke from the teacher.
     
  4. Dec 17, 2004 #3

    quasar987

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    Yes, I transcripted the question as it was written on the paper!

    And the worst part is that a friend of mine went to see a teacher about that question today and he told him the way to do it was to use the limit to show that the gradient was (0,0) at (0,0) !! Insane! :yuck:

    The exam is at: http://www.dms.umontreal.ca/~giroux/documentsc1/finalC1H04.html and it is the first question. Although it is in french, you can see that the "including at (0,0)" part is there.
     
    Last edited: Dec 17, 2004
  5. Dec 17, 2004 #4

    dextercioby

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    Il n'y a acune problème.Je comprends Français.
    Yes,i saw that.He meant to 'compute that limit'.It's (0,0),as it can easily be shown.However,it is still weird.He could have said:"Find the limit."Students would have gotten it.

    Daniel.

    PS.Je pense que le proffesseur Giroux (?) est un idiot. :tongue2:
     
  6. Dec 18, 2004 #5

    Integral

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    Oh My, Oh My!

    you must keep in mind that this is function of 2 variables, You need to be a bit more careful.

    First note that if you plot points on either side of zero the function is well behaved.

    The correct way to analysis multi variable limits is to hold one variable constant, while the limit in the other variable is examined.

    so if you first evaluate your partial of f wrt x at y=0, you have

    [tex]\left[ \frac{\partial f}{\partial x} \right]_{(x,0)} = \left[\frac{4x^3}{x^2}-\frac{2x^5}{(x^2)^2}-\frac{0}{(x^2)^2}\right]_{(x,0)}=2x [/tex]

    Clearly this is 0 at x = 0

    You will get similar results if you hold x constant (that is compute the partial wrt y) and take the y limit as x is held constant.

    You will find that the function is well behaved and approaching zero for all x,y near zero. therefore the limit exists at (0.0) and is 0. This can also be seen by plotting points near zero.
     
  7. Dec 18, 2004 #6

    quasar987

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    But do we agree that finding the limit of the gradient as (x,y) approches (0,0) is not the same as "calculating" the gradient for the value (0,0) ?

    Are you saying it is not necessary to use the epsilon-delta definition to show that the limit is 0?
     
  8. Dec 18, 2004 #7

    arildno

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    1. Prove that
    [tex]\lim_{(x,y)\to(0,0)}f(x,y)=0[/tex]
    We must show that given [tex]\epsilon>0[/tex] there exist a [tex]\delta>0[/tex]
    so that [tex]||\vec{x}-\vec{0}||<\delta\to|f(\vec{x})-0|<\epsilon, \vec{x}\equiv(x,y)[/tex]
    Rewrite your function in polar form:
    [tex]x=r\cos\theta,y=r\sin\theta\to{f}(x,y)=g(r,\theta)=r^{2}\cos^{4}\theta+r\sin^{3}\theta[/tex]
    Hence, since:
    [tex]|g(r,\theta)-0|<=r^{2}+r[/tex]
    it follows that we can find, for [tex]\epsilon>0[/tex] a [tex]\delta>0[/tex] so that
    [tex]r<\delta\to{r}^{2}+r<\epsilon[/tex]
    [tex]\delta<min(\sqrt{\frac{\epsilon}{2}},\frac{\epsilon}{2})[/tex]
    is perhaps the simplest choice.

    Hence, defining f(0,0)=0 gives us a continous function.
    Was this part of what you asked for?
     
  9. Dec 18, 2004 #8

    arildno

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    2. By definition,
    [tex]\frac{\partial{f}}{\partial{x}}\mid_{(x_{0},y_{0})}=\lim_{x\to{x}_{0}}\frac{f(x,y_{0})-f(x_{0},y_{0})}{x-x_{0}}[/tex]
    Hence, in your case:
    [tex]f(x,0)=x^{2},f(0,0)=0,f(0,y)=y[/tex]
    We therefore gain:
    [tex]\frac{\partial{f}}{\partial{x}}\mid_{0,0}=\lim_{x\to0}x=0,\frac{\partial{f}}{\partial{y}}\mid_{0,0}=\lim_{y\to0}1=1[/tex]

    Hmm..I'll go and check up the definition to be sure, but I'm pretty certain I'm right here..
     
    Last edited: Dec 18, 2004
  10. Dec 18, 2004 #9

    arildno

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    3. It is NOT correct that, for example [tex]\frac{\partial{f}}{\partial{x}}[/tex] is continuous at (0,0):
    We compute [tex]\frac{\partial{f}}{\partial{x}}[/tex] at [tex](x_{0},y_{0})\neq(0,0)[/tex]
    [tex]\frac{\partial{f}}{\partial{x}}\mid_{(x_{0},y_{0})}=\frac{4x_{0}^{3}(x_{0}^{2}+y_{0}^{2})-2x_{0}(x_{0}^{4}+y_{0}^{3})}{(x_{0}^{2}+y_{0}^{2})^{2}}[/tex]

    Rewriting in polar form:
    [tex]\frac{\partial{f}}{\partial{x}}=h(r,\theta),x_{0}=r\cos\theta,y_{0}=r\sin\theta[/tex]
    We get:
    [tex]h(r,\theta)=r(4\cos^{3}\theta-2\cos^{5}\theta)-2\cos\theta\sin^{3}\theta[/tex]

    Hence,
    [tex]\lim_{(x_{0},y_{0}\to(0,0)}\frac{\partial{f}}{\partial{x}}\mid_{(x_{0},y_{0})}[/tex]
    depends on the angle by which we approach the origin.

    In particular, the limit value will not always equal zero
    ,which is the function value [tex]\frac{\partial{f}}{\partial{x}}\mid_{(0,0)}[/tex]
    Hence the x-derivative is discontinuous at the origin.
     
    Last edited: Dec 18, 2004
  11. Dec 18, 2004 #10

    arildno

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    4. It is instructive to compute the DIRECTIONAL DERIVATIVES at the origin.
    by definition, the derivative in the direction [tex]\vec{n},||\vec{n}||=1[/tex]
    is given by:
    [tex]\frac{\partial{f}}{\partial{n}}\mid_{(x_{0},y_{0})}\equiv\lim_{t\to0}\frac{f(x_{0}+tn_{x},y_{0}+tn_{y})-f(x_{0},y_{0})}{t}[/tex]
    Setting [tex]\vec{n}=n_{x}\vec{i}+n_{y}\vec{j}=\cos\theta\vec{i}+\sin\theta\vec{j}[/tex]
    we gain:
    [tex]\frac{\partial{f}}{\partial{n}}\mid_{(0,0)}=\sin^{3}\theta[/tex]
     
  12. Dec 18, 2004 #11
    Hmm...mais vous n'écrivez pas si bien... :wink:

    I kidd.
     
  13. Dec 19, 2004 #12

    dextercioby

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    Oui,c'est vrai.J'ai oublié un "u"... :confused: "aucune",peut [itex]\hat{e}[/itex]tre...??

    Daniel.
     
  14. Dec 19, 2004 #13

    quasar987

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    Thanks arildno. I don't think this is what the question asked but they were assuredly very instructive posts, especially #4.
     
  15. Dec 19, 2004 #14

    arildno

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    Just a final comment.
    This function is an example where the partial derivatives all exist at (0,0), but still is not a differentiable function at (0,0).
    If you want "gradient" to mean the "(total) derivative of f", then the gradient
    doesn't exist at (0,0).
     
  16. Dec 19, 2004 #15

    mathwonk

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    By definition, the total derivative of f(x,y) equals 0 at (0,0) if and only if

    f(x,y)/|(x,y)| goes to zero as (x,y) does, where |(x,y)| = sqrt(x^2+y^2). in the case above f(x,y)/|(x,y)| = {x^4 + y^3}/(x^2+y^2)^(3/2).

    thus the bottom has homogeneous degree 3, while the top has partly degree 4 and partly degree 3. thus if you approach (0,0) along the line x=0, this quotient is equal to 1, hence does not approach 0.

    can you see how to modify the top of f slightly so as to get a function which does have total derivative zero at (0,0)? and by the way the professor is not an idiot here. indeed this is the kind of question which exposes whether one does or does not know what a derivative means.
     
  17. Dec 20, 2004 #16

    arildno

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    Just adding a post-ultimate comment:
    In order to prove that f is not differentiable at zero (that is, that the "total derivative" doesn't exist), you must show that there cannot exist a vector (a,b) independent of [tex]\vec{x}=(x,y)[/tex] so that:
    [tex]\lim_{(\vec{x})\to(0,0)}\frac{|f(x,y)-f(0,0)-ax-by|}{||\vec{x}||}=0[/tex]

    As mathwonk proved, (a,b)=(0,0) cannot be used, and it is a simple matter to show that no other choice works either.

    (The simplest way to do this, is to change (x,y) into polar coordinates and show that (a,b) must be a function of the angle of approach in order to get zero (that is, dependence on (x,y)))

    If you could have found a vector (a,b), that would, of course, have been the "total derivative", "gradient" of f at (0,0)
     
    Last edited: Dec 20, 2004
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