How to show that $\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^3}$ does not exist? I have a question. I have to check whether the limit exist. But I want to check it with a) two paths or b) polar coordinates, but it doesn't work. It is about  $\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^3}$. Can anyone help with to show that the limit does not exist?
Thank you
 A: Let $f(x,y) := \frac{xy^2}{x^2+y^3}$. Then we have
$$f(x,y) = 1 \Leftrightarrow xy^2 = x^2+y^3 \Leftarrow x=\underbrace{\frac{y^2+\sqrt{y^4-4y^3}}2}_{=:h(y)}$$
Let $y_n:=1/n, x_n:=h(y_n)$. Then $f(x_n, y_n)\equiv 1$ and $$\lim_{n\to\infty}x_n = \lim_{n\to\infty}h(y_n) = h(\lim_{n\to\infty}x_n)= h(0) = 0\text{,}$$since $h:(-\infty,4)\to \mathbb R$ is continuous. Thus, $(x_n,y_n)\to_{n\to\infty}(0,0)$.
Also $f(0, 1/n) \equiv 0$.
So we have $\lim_{n\to\infty} f(x_n, y_n)\neq\lim_{n\to\infty} f(0, 1/n)$, and thus $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.
A: I know this answer is late. But for future interest, someone may be interest in a method that can be applied for disproving the existence of limits where we have rational functions. However, I don't know to what extent this method is applicable.

First step is to realize what set of $(x,y)$ that we're trying to avoid. Let's define $M := \{(x,y) : x = t^{1/2}, y = -t^{1/3} \}$. These values for $(x,y)$ gives us a $0$ in the denominator, hence needs to be avoided.

Now that's the first step done. Now, we pick our $x$ from our set $M$ such that it's relatively close to $t^{1/2}$. Let $x = (1+\epsilon(t))^{1/2}t^{1/2}$ where $\epsilon(t) \neq 0 \forall t \neq 0$. For now, we won't look much into our $\epsilon(t)$ function. Furthermore, let $y = -t^{1/3}$ as given in our set $M$.

Substituting this into our original expression, we have that:
$$\lim_{t\rightarrow 0} \frac{t^{1/2} \cdot t^{2/3} \cdot (1+\epsilon(t))^{1/2}}{t\cdot\epsilon(t)}$$
Simplifying and defining, we get:
$$L_{\epsilon} := \lim_{t\rightarrow 0} \frac{t^{1/6} (1+\epsilon(t))^{1/2}}{\epsilon(t)}$$
Now, it becomes trivial in choosing our $\epsilon(t)$ to form a contradiction.
Let $\epsilon(t) = 1$, we get that $L_{\epsilon} = 0$, but choosing $\epsilon(t) = t^{1/6}$, we instead get that $L_{\epsilon} = 1$.
Hence, since two different paths for $(x,y)$ yield different limits, it's nonexistent.

Feel free to add any comments for improvments or any missing details.
