Bounded parts of a function in limit study I'm studying limits in functions with 2 independent variables. I have this solved limit:
$$\lim\limits_{(x,y) \to (0,0)} \frac{-3x^2y}{x^2+y^2} = \lim\limits_{(x,y) \to (0,0)} -3y\frac{x^2}{x^2+y^2} = 0\cdot bounded=0$$
With that in mind, why is this not bounded?
$$\lim\limits_{(x,y) \to (0,0)} \frac{-3xy}{x^2+y^2} = \lim\limits_{(x,y) \to (0,0)} -3y\frac{x}{x^2+y^2} = \dots$$
Have in mind that I'm not a native English speaker, in case "bound" is not the correct word I hope you get the right idea of what I am asking. Apologies.
 A: First of all you are using the word 'bounded' correctly. For what you mean is that the limit is maintained in a controlled region of values such that the other limit is determinant for the behavior of the entire limit. Second, you can see that
$$\left \vert \frac{x^2}{x^2+y^2} \right\vert \leq 1$$ 
Then you can use the squeeze theorem
\begin{equation}
-1\left \vert -3y\right \vert \leq \frac{-3yx^2}{x^2+y^2} \leq \vert -3y \vert
\end{equation}
and this is true for every $(x,y) \neq (0,0)$ so, you can calculate the limit 
$$\lim_{(x,y)\rightarrow(0,0)}\vert -3y \vert = 0$$
So you will have the answer that you wrote above. But now you have to note that the $ x/(x^2+y^2)$ is not an bounded function. Calculate the limit for different continuous curves and you will see that these limits are different, so the limit cannot exist. Pic for example the $\gamma_1(t) = (0,t)$ and $\gamma_2(t) = (t,t)$ so you will have, defining $f(x,y) = -3xy/(x^2+y^2)$
$$\lim_{t \rightarrow 0}f(\gamma_1(t)) = \lim_{t \rightarrow 0}\frac{-3t0}{0^2+t^2} = 0$$ and 
$$\lim_{t \rightarrow 0}f(\gamma_2(t)) = \lim_{t \rightarrow 0}\frac{-3t^2}{t^2+t^2} = \frac{-3}{2}$$
Because these limit's are different the limit does not exist. 
