I am aware there are many questions similar to this on MSE but I am having trouble following any of the solutions given.

I have the function given by

$f(x) = \dfrac{xy}{x^2+y^2}\;$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$

I have calculated the partial derivatives and found:

$$\frac{\partial f}{\partial x} = \frac{y(-x^2+y^2)}{(x^2+y^2)^2}, \quad \frac{\partial f}{\partial y} = \frac{x(-y^2+x^2)}{(y^2+x^2)^2}$$

Now I need to show the partial derivatives exist for $(x,y) = (0,0)$. I am also asked to show it is not continuous at $(x,y)=(0,0)$, based on other answers I've seen it seems like this follows from solving the first part but I fail to see how that follows through as well.

I could just copy the solutions with my function in place as I have seen a lot of answers using the definition of the derivative, but realistically I do want to understand what the thinking behind this is.

Any help would be appreciated.

  • 1
    $\begingroup$ Since $xy/(x^2+y^2)$ is undefined at the origin, you have to go back to the basic limit definitions of the partial derivatives. $\endgroup$ – amd Mar 25 '17 at 18:18

Why we can use the definition is pretty obvious - if something does not satisfy the defining criteria of some property then that thing does not possess the property - this is the nature of definitions. Why we should use the definition, well who's to say. If consideration of a definition leads to an easy answer to a question we have, why not consider it?

Consider the quotient definition for partial differentiation: let $\mathbf{0}=(0,0)$ and $\mathbf{e}_1,\mathbf{e}_2$ be the standard basis for the plane. For any nonzero $t \in \mathbb{R}$ we have

$$Q(t) := \frac{f(\mathbf{0} + t\mathbf{e_1}) - f(\mathbf{0})}{t} = \frac{f(t,0) - f(0,0)}{t} = 0.$$

This is true because $f(t,0) = f(0,0) = 0$ for any such $t$. By definition $$\frac{ \partial f}{ \partial x}(\mathbf{0}) := \lim_{t\to 0} Q(t) = 0.$$

Similarly, $$\frac{ \partial f}{\partial y}(\mathbf{0}) := \lim_{t\to 0} \frac{f(\mathbf{0}+ t\mathbf{e}_2) - f(\mathbf{0})}{t} = \lim_{t\to 0}\frac{f(0,t) - f(0,0)}{t} $$ $$= \lim_{t \to 0}\frac{0 - 0}{t} =\lim_{t \to 0}\frac{0}{t} = \lim_{t\to 0} 0 = 0. $$

As for continuity of $f$ at the origin, if it were true that $f$ is continuous there, then for any sequence $\mathbf{x}_n \to \mathbf{0}$ we would have $\lim_{n \to \infty} f(\mathbf{x}_n) = f(\mathbf{0}) = 0$. But the sequence $\mathbf{x}_n = \left(\frac{1}{n},\frac{1}{n}\right) \to \mathbf{0}$ is such that $f(\mathbf{x}_n) = 1/2$ for every index $n$. It certainly isn't true then that $f(\mathbf{x}_n) \to \mathbf{0}$, and we deduce that $f$ cannot be continuous at the origin.

  • $\begingroup$ $Q(t)$ is just part of the definition - note that it is constructed using the original function $f$, the specific point $\mathbf{0}$ (which is fixed in the definition of $Q$) and also on the directional vector $\mathbf{e}_1$ (which is also held fixed). By definition the partial $\frac{ \partial f}{\partial x}(\mathbf{0})$ exists if $Q(t)$ has a limit $\ell$ as $t \to 0$, and in this case $\frac{ \partial f}{\partial x}(\mathbf{0})$ is set equal to $\ell$. $\endgroup$ – joeb Mar 25 '17 at 18:55
  • $\begingroup$ Ah! I just understood it and deleted my comment before seeing your response, thank you! :) $\endgroup$ – Evan Mar 25 '17 at 18:56

The two questions about partial derivatives and continuity are independent. For partial derivatives, you can use the definition, as in other answer, but you can also think geometrically.

The partial derivative $\frac{\partial f}{\partial x}$ at $(0,0)$ is the derivative of the curve that results from the intersection of the surface and the vertical plane $y=0$. In other words, perform the substitution $y=0$ and you obtain the 1-dimensional function: $g(x)=f(x,0)=0$. Therefore, $$\frac{\partial f}{\partial x}(0,0)=g'(0)$$ And, yes, the function $g$ seems pretty derivable.

As for continuity, the limit does not exist, e.g. by comparing trajectories $x=0$ and $x=y$.

  • $\begingroup$ What do you mean by comparing trajectories? I understand that you have to approach the limit from each side and then conclude they are not equal hence the limit does not exist and it is not continuous. But in the practical sense, to show this do I apply the definition to $f(x,0)$ and $f(x,x)$ ? $\endgroup$ – Evan Mar 25 '17 at 19:11
  • $\begingroup$ @N.K Unlike the situation for functions of a single variable, it’s not enough to approach from “each side” (how many sides is that, anyway)? For a limit to exist, you have to get the same value for all ways to approach the point. For example, there are functions for which you get the same limiting value along any straight line, but not along a curved path. $\endgroup$ – amd Mar 25 '17 at 20:30
  • $\begingroup$ @N.K I am not sure I have interpreted right your question. Do you mean you have to prove that the partial derivatives are not continuous? My answer refers to proving that the function is discontinuous. $\endgroup$ – Miguel Mar 26 '17 at 15:28
  • $\begingroup$ @amd But it is enough to compare two trajectories to prove that the limit does NOT exist (if you are lucky enough to find two trajectories that differ) $\endgroup$ – Miguel Mar 29 '17 at 10:35

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