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 $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0?$$

Just notice that since $\alpha > 0$ and $x^2 \geq 0$ you have that
$$ \frac{x^\alpha y^4}{x^2+y^4} \leq \frac{x^\alpha y^4}{y^4} = x^\alpha \overset{(x,y) \to 0}{\longrightarrow} 0.$$
Thus it is clear that one has $\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4} = 0$. You rarely need polar coordinates for these kinds of questions. The most problems I encountered personally are easy to solve by using this trick above. I hope it helps you :)
