continuity of a multivariable function at a point

Question

I want to know whether the function $\dfrac{x^2y}{x-y}$ is continuous at $(0,0)$ or not. when I switch to polar coordinate i get the limit of : $\dfrac{(r \cos^2 \theta \sin \theta)}{\cos \theta - \sin \theta}$ when r is going to zero. can I say that this limit does not exist because there are points where $\theta = \dfrac{\pi}{4}$ and then the denominator is zero, and therefore the function is not continuous at $(0,0)$ ? the value of the function when x is equal to y is 0 thank you

Answer 1

Since$$\lim_{n\to\infty}f\left(\frac1n,\frac1n+\frac1{n^4}\right)=\lim_{n\to\infty}-\frac1{n^2}-n=-\infty,$$then, no matter how you define $f(0,0)$, the function $f$ will always be discontinuous at $(0,0)$.

Answer 2

Let set $\ u=x-y\ $ then $$f(x,y)=\dfrac{x^2y}{x-y}=\dfrac{x^2(x-u)}{u}=\dfrac{x^3}u-\underbrace{x^2}_{\to 0}$$

So the behaviour in $(0,0)$ is determined by the relative "force" of $x$ and $u$.

This decomposition helps you finding counter examples.

For instance if $u\sim x^a$ then $\dfrac{x^3}{u}\sim x^{3-a}$

For $a>3$ this quantity tends to infinity, and we cannot extend by continuity.

• this is J.C.Santos example above with $a=4$, let's take $x=\frac 1n$ then $y=\frac 1n-\frac 1{n^4}$

For $a=3$, you get a constant term, and this time we cannot extend by continuity because we can exhibit multiple limits.

• indeed $f(\frac 1n,\frac 1n-\frac 1{\ell\ n^3})\to \ell\quad$ for any $\ell$ we want.