Limit of a function with 2 variables I am given this function: $$f(x,y)=\begin{cases}\frac{xy^3}{x^2+y^4} & \text{ for } (x,y)\not=(0,0)\\ 0 & \text{ for } (x,y)=(0,0)\end{cases}$$ and I have to check if it is continuous in $(0,0)$. Therefore I want to calculate $\lim \limits_{(x,y) \rightarrow 0}{\frac{xy^3}{x^2+y^4}}$.   
I already tried substituting and polar coordinates but did not come to a solution yet.
Can someone give a few possibilities to calculate a limit of a function with multiple variables. Which is the best one to try first? I found that polarcoordinates often works very well.
There has already been a question concerning this function and it's partial derivates and if it's differentiable. You may find that one here:   determine whether $f(x, y) = \frac{xy^3}{x^2 + y^4}$ is differentiable at $(0, 0)$.
Thanks!
 A: Approach $(0,0)$ from a few different paths, and you will find that it appears the limit is in fact $0$. To prove this is the case, you can use the Squeeze Theorem. 
We have that 
\begin{align*}
\bigg|\frac{xy^{3}}{x^{2} + y^{4}} - 0 \bigg| &\leq \bigg|\frac{xy^{3}}{2xy^{2}}\bigg| && \text{using the inequality } 2ab \leq a^{2} + b^{2}
\end{align*}
and from here it should be easy to find a bounding function and show that its limit is indeed $0$, which implies by the Squeeze Theorem that the limit of $f$ as $(x,y) \rightarrow (0,0)$ is $0$ as well.  
A: $$\lim\limits_{(x,y) \to (0,0)}{\frac{xy^3}{x^2+y^4}}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+}{\frac{r^4\cos\phi \sin^3\phi}{r^2\cos^2\phi+r^4\sin^4\phi}}$$
$$=\lim\limits_{r\to 0^+}{\frac{\cos\phi \sin^3\phi}{\frac{\cos^2\phi}{r^2}+\sin^4\phi}}$$
$$={\frac{\cos\phi \sin^3\phi}{\lim\limits_{r\to 0^+}\frac{\cos^2\phi}{r^2}+\sin^4\phi}}$$
$$={\frac{\cos\phi \sin^3\phi}{\cos^2\phi\lim\limits_{r\to 0^+}\frac{1}{r^2}+\sin^4\phi}}$$
Note that
$$\lim\limits_{r\to 0^+}\frac{1}{r^2}=\infty$$
Therefore
$${\frac{\cos\phi \sin^3\phi}{\cos^2\phi\lim\limits_{r\to 0^+}\frac{1}{r^2}+\sin^4\phi}}=0$$
And
$$\lim\limits_{(x,y) \to (0,0)}{\frac{xy^3}{x^2+y^4}}=0$$
