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.
Best Answer
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.