Find $\lim_\limits{(x,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}$ where $\alpha > 0$

Question

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

Answer 1

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 :)

Answer 2

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.