Multivariable function limit How to approach this: $\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y}$?
Been able to grind $\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^2}$, it is in(link) finnish, but formulas and idea should be selfevident. However $\dot +y$ instead of $\dots+y^2$ in divisor confuses me.
Or am I thinking in completely wrong direction?
 A: First of all, you need to consider your space restricted to $\mathbb{R}^2 - \{(x,y):x^2+y=0\}$. Otherwise you cannot form a neighbourhood of the origin where the function is defined.
Even then, consider the trajectories $x=0$ and $x^2=y^2-y$:
$$\lim_{x=0,y\to 0} \frac{x^2 y}{x^2+y} = \lim_{y\to 0} \frac{0}{y}=0$$
$$\lim_{x^2=y^2-y,y\to 0} \frac{x^2 y}{x^2+y} = \lim_{y\to 0} \frac{y^3-y^2}{y^2}=-1$$
Since the limit differs along the two trajectories, then the limit does not exist.
The morale is: it is tricky and requires some experience to find a suitable trajectory, but at least you need to have the intuition that the limit does not exist. A useful observation is the presence of odd powers.
A: $f(x,y)=\frac{x^2y}{x^2+y}$
Suppose the limit exist and it is finite. Then, let's take $\epsilon > 0$. There is $\delta > 0$ so that $|\frac{x^2y}{x^2+y} - L|<\epsilon$ for all $x,y$ so that $x^2 + y^2 < \delta ^2$. 
Let's consider $x_n= \frac{1}{\sqrt n}, y_n= -\frac{1}{n + 1}$. There is N so that $x_n^2 + y_n^2<\delta^2, \forall n > N$. We have $f(x_n,y_n)=-1$.
So $|-1 - L| < \epsilon$ for arbitrary $\epsilon$, therefore $L = -1$.
Now, consider $x_n= \frac{1}{\sqrt n}, y_n= -\frac{1}{n + 2}$. By following the same steps as above, we'll get $L = -\frac{1}{2}$.
It is proven that there is no finite limit. Similar for infinite limit 
A: Here's another approach
$$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+}\frac{r^2\cos^2\phi\sin\phi}{r\cos^2\phi+\sin\phi}$$
Now lets attempt to find bounds that are independent of $\phi$
$$\frac{r^2\left|\cos^2\phi\sin\phi\right|}{\left|r\cos^2\phi+\sin\phi\right|}\leq\frac{r^2}{\left|r\cos^2\phi+\sin\phi\right|}$$
Since this limit is dependent on $\phi$, we can conclude that
$$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y}=\mbox{non existent}$$
