Find $\lim_\limits{(x,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}$ where $\alpha > 0$ How can we find the following limit? $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}\qquad \alpha>0$$
By using the polar coordinate, we get
$$\lim_{r\to 0}r^{\alpha+2}\frac{\cos^\alpha\theta \sin^4\theta}{\cos^2\theta+r^2\sin^4{\theta}}=0$$ if $\theta\notin\{\frac{\pi}{2}+\pi k:k\in\Bbb Z\}$.
Now, if $\theta = \frac{\pi}{2}+\pi k$ for some $k\in\Bbb Z$, then we get $$\lim_{(0,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0.$$
Can we conclude that $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0?$$

From the AM-GM inequality we have $$\left|\frac{x^ay^b}{x^2+y^4}\right|\le \frac12 |x|^{a-1}|y|^{b-2}$$
where we assume that either $a$ is such that $x^a\in \mathbb{R}$ for $x$ in a neighborhood of $0$ or that the limit is taken as $(x,y)\to (0^+,0)$.  
And equality holds when $x=\pm y^2$.  Hence, we find 
$$\lim_{(x,y)\to(0,0)}\frac{x^ay^b}{x^2+y^4}=0$$
whenever we have $2a+b>4$. The limit fails to exist otherwise.
In the case at hand, $a=\alpha$ and $b=4$ and we see that the limit is $0$ for $\alpha>0$ and fails to exist otherwise.
