For which $a\in \mathbb{R}$ is $f(x,y) = \frac{xy^a}{x^2+y^2}$ with $f(0,0)=0$ continuous?

Question

My function is defined as: $$f(x,y) = \begin{cases} \dfrac{xy^a}{x^2+y^2}&(x,y) \neq (0,0)\\ \ 0&(x,y) = (0,0) \end{cases} $$

I want to find out for which $a\in \mathbb{R}$ the function is continuous. I am not quite sure how to solve something like this. I know that for $a=1$ the function is definitely not continuous, which can be checked, since $(\frac1n,\frac1n)$ converges to $(0,0)$ but $f(\frac1n,\frac1n)$ does not. Now if I want $f$ to be steady in $(0,0)$, I figured that this is the case if $a>1$. But this is where I get stuck. How to I prove or check, if for $a>1$ the rest of the function is continuous as well? Any help is greatly appreciated!

Answer 1

Exercise 1.

Note that if $x=r\cos\theta$ and $y=r\sin\theta$ $$ \left|\frac{xy^a}{x^2+y^2}\right|=\frac{r^{1+a}}{r^2}\vert\cos\theta\sin^a\theta\vert\leq r^{a-1}. $$ which gives you a sufficient condition that $f$ is continuous at $(0,0)$: $a>1$.

On the other hand, you want also a necessary condition.

Exercise 2.

Show that when $a\leq 1$, $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.

Again, consider $x=r\cos\theta$, $y=r\sin\theta$ and $$ \frac{xy^a}{x^2+y^2}=r^{a-1}\cos\theta\sin^a\theta. $$ If $a<1$, then $f$ is not even bounded near $(0,0)$, which tells (why?) you that $f$ cannot be continuous at $(0,0)$. If $a=1$, then $$ \frac{xy^a}{x^2+y^2}=\cos\theta\sin^a\theta $$ which depends on $\theta$, this implies (why?) that $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.

Answer 2

$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{r\to0}f(r\cos\theta,r\sin\theta)=\lim_{r\to 0}r^{a-1}\cos\theta\sin^a\theta$$ Note that the limit equals $0$ if it exists, and if limit equals $0$ implies $f$ is continuous.
The limit does not exist implies that $f$ is not continuous at the origin. And for $a\le 1$, the limit does not exist

So $f$ is continuosu for $a>1$

Answer 3

There are lots of tricks to these types of limit problems. Solving them seems a bit ad hoc/heuristic at times. You are correct, for $a=1$ this is not continuous at $(x,y)=(0,0)$. In general you want to check that the exponents in the numerator and denominator don't add to the same integer value for $a$. Note that for $a=1$ $lim_{x,y\rightarrow (0,0)}$ $\frac{xy}{x^2+y^2}$ does not exist along the path $y=x$ because we get that the limit is equal to $1/2$. If we have $a\geq2$ this cannot happen, and the function will be continuous at the origin if we take the limit along any path heading towards $(0,0)$. You can save yourself algebra this way as well rather than finding convergent sequences like ${1/n}$ for the coordinates $(x,y)$. Also, you can try to use polar coordinates as the answers above indicated by simply substituting $x=rcos(\theta)$ and $y=rsin(\theta)$.

Answer 4

The point where you may have some problems is the origin right?

So let's see what happens when we take the limit in polar coordinates: $$ \lim_{r\rightarrow 0} \frac{r\cos t r^a\sin^a t}{r^2}= \lim_{r\rightarrow 0} \frac{r^{a+1}\cos t \sin^a t}{r^2} $$ Which you need to tend to 0 independently of $t$. From this, it can be seen that you must have $a+1>2\Rightarrow a>1$