Given the function $f(x,y)=\frac{5xy^2}{x^2+y^2}$, argue analytically, if the discontinuity of $f$ at $(0,0)$ is avoidable or unavoidable.

Question

It is clear that $f(0,0)$ does not exist, and therefore the function is discontinuous. Now to know if the discontinuity is avoidable or inevitable, we must see that the limit at that point exists or not. To see the above we will use polars. \begin{align*} \lim_{(x,y)\rightarrow(0,0)}\frac{5xy^2}{x^2+y^2}&=\lim_{r\rightarrow0}\frac{5(r\cos(\theta))(r\sin(\theta))^2}{(r\cos(\theta))^2+(r\sin(\theta))^2}\\ & =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2(\cos^2(\theta)+\sin^2(\theta))}\\ & =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2}\\ & =\lim_{r\rightarrow0}5r\cos(\theta)\sin^2(\theta)=0. \end{align*} Now let's test for the definition $\epsilon-\delta$, to see that the limit exists and is zero. \begin{align*} \left|\frac{5xy^2}{x^2+y^2}-0\right| & =\left|\frac{5xy^2}{x^2+y^2}\right|\\ & =\frac{5|x|y^2}{x^2+y^2}\\ & \leq 5|x| \text{ since $y^2\leq x^2+y^2$}\\ & \leq 5\sqrt{x^2+y^2} \text{ since $|x|\leq\sqrt{x^2+y^2}$}\\ & <5\delta = \epsilon. \end{align*} Therefore we have: \begin{align*} \forall\epsilon>0, \exists\delta>0 \text{ such that, if } 0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-0| &\leq5 \sqrt{x^2+y^2}\\ &<5 \delta = 5\left(\frac{\epsilon}{5}\right)=\epsilon. \end{align*} Thus the limit exists and is zero, therefore the discontinuity of $f$ is avoidable.

I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.

Answer 1

Both your proofs (the analytic proof, and the proof using polar coordinates) that $\lim_{(x,y) f(x,y)\rightarrow (0,0)}=0$ are correct.

Answer 2

Your first sentence is wrong; since $f$ is undefined at $(0,0)$, it is neither continuous nor discontinuous at that point.

Otherwise, you proved correctly that $\lim_{(x,y)\to(0,0)}f(x,y)=0$ and that therefore you can extend $f$ in such a way that it becomes continuous at $(0,0)$.