I am trying to find whether or not the limit
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^2}{x^5+y^3},$$ $$x^5+y^3\neq0$$
exists. According to my teacher the limit is not supposed to exist, and here comes the part where I am stuck.
I did try to use polar coordinates $$x=rcos(\theta)$$ $$y=rsin(\theta)$$ and it got me to $$\frac{x^3y^2}{x^5+y^3}=\frac{r^3cos^3(\theta)r^2sin^2(\theta)}{r^5cos^5(\theta)+r^3sin^3(\theta)}=\frac{r^2cos^3(\theta)sin^2(\theta)}{r^2cos^5(\theta)+sin^3(\theta)}$$
How can I proceed from here to prove that the limit does not exist ? Every time I try to substitute something it always gives me the same answer, which is that if the limit exists, it is equal to $0$.
I link to a handwritten solution to the problem below that I can't seem to understand, so if anyone is willing to explain what my teacher is actually doing I would appreciate it greatly!
(There are a few words in swedish that I don't think you need to translate to be able to understand the actual solution)
The clue here is to approach on a path very close to $x^5+y^3=0$. Your teacher suggests noticing that you can parametrize this curve by $$x=t^{1/5}, y=-t^{1/3}.$$ A nearby path is obtained by multiplying $y$ by $(1+\epsilon(t))^{1/3}$, where $\epsilon(t)\to 0$ as $t\to 0$. Taking $\epsilon(t)=t$ shouldn't be too difficult. What happens, then, to $f(x,y)$?
We have $$f(t^{1/5},-t^{1/3}(1+t)^{1/3}) = \frac{t^{3/5}t^{2/3}(1+t)^{2/3}}{t-t(1+t)} = \frac{t^{19/15}(1+t)^{2/3}}{-t^2} = -\frac{(1+t)^{2/3}}{t^{11/15}},$$ and the absolute value of this goes to $\infty$.