Question

Check if limit exists at (x,y)=(0,0).
f(x,y) = (x^3y)/(x^4+y^2) .

Answer 1

What have you tried?

Checking numerically, the limit does seem to exist, but I'm wondering if there's a convenient analytic approach.

Answer 2

Converting to polar may help:
\[\begin{cases}x=r\cos t\\y=r\sin t\end{cases}~~\implies~~\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}=\lim_{r\to0^+}\frac{r^4\cos^3t\sin t}{r^2\left(r^2\cos^4t+\sin^2t\right)}\]
One can show that \(|\cos^3t\sin t|\le\dfrac{3\sqrt3}{16}\). You also have that
\[|r^2\cos^4t+\sin^2t|\le r^2|\cos^4t|+|\sin^2t|\le r^2+1\]
By the squeeze theorem, then, you have
\[\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}=\lim_{r\to0^+}\frac{r^4\cos^3t\sin t}{r^2\left(r^2\cos^4t+\sin^2t\right)}=\frac{3\sqrt3}{16}\lim_{r\to0^+}\frac{r^4}{r^4+r^2}\]