$$\lim_{(x,y)\to (0,0)}\frac{x^3+y^4}{x^2+y^2}$$
What I have manage to come up with is:
$$\lim_{(x,y)\to (0,0)}\left|\frac{x^3+y^4}{x^2+y^2}\right|\leq \lim_{(x,y)\to (0,0)}\left|\frac{x^3+y^4}{2xy}\right|\leq \lim_{(x,y)\to (0,0)}\left|\frac{x^4+y^4}{2xy}\right|$$
taking $x=r\cos\theta$ and $y=r\sin\theta$
we get $$\lim_{r\to 0}\left|\frac{r^4(\cos^4\theta+\sin^4\theta)}{2r^2\cos\theta\sin\theta}\right|=\lim_{r\to 0}\left|\frac{r^4}{2}\cdot\frac{\cos^4\theta+\sin^4\theta}{2\cos\theta\sin\theta}\right|$$ which is a bounded function and a function going to zero, and there for the limit is zero and by the squeeze limit
$$\lim_{(x,y)\to (0,0)}\frac{x^3+y^4}{x^2+y^2}=0$$
Is the way is valid? is there a simpler way?