# Determine $p$ for which the limit $\lim \limits_{(x,y) \to (0,0)} \frac{|x||y|^{p}}{x^2+y^2}$ exists.

$$\lim \limits_{(x,y) \to (0,0)} \frac{|x||y|^{p}}{x^2+y^2}$$

Using manipulation of powers, $|x| = (|x|^2)^{\frac{1}{2}}$ and $|y| = (|y|^2)^{\frac{p}{2}}$

this is the inequality I've arrived at:

$$\frac{|x||y|^{p}}{x^2+y^2} \leq \frac{(|x|^2+|y|^2)^{\frac{1}{2}}(|x|^2+|y|^2)^{\frac{p}{2}}}{x^2+y^2}$$

Since they're square numbers, we can disregard the absolute values and get

$$(x^2+y^2)^{\frac{p}{2}-\frac{1}{2}}$$

Which is only a good "squeeze" function if $\frac{p}{2} - \frac{1}{2} > 0$

Hence for the limit to exist $$p>1$$

Would this be complete?

If so, what about values of p for which the limit doesn't exist?

Is it simply $p\leq1$?

You can use polar coordinates $x = r\cos\phi$, $y = r\sin\phi$.

$$\lim \limits_{(x,y) \to (0,0)} \frac{|x||y|^{p}}{x^2+y^2} = \lim_{r\to 0}\frac{r\left|\cos\phi\right|r^p\left|\sin\phi\right|^p}{r^2} = \lim_{r\to 0} r^{p-1}\left|\cos\phi\right|\left|\sin\phi\right|^p$$

If $p - 1 > 0 \iff p > 1$ then this clearly converges to $0$.

If $p = 1$, then the limit converges to $\left|\cos\phi\sin\phi\right| = \frac{1}{2}\left|\sin 2\phi\right|$, which can assume different values as we approach the origin along rays with different angles $\phi$. Hence, the limit doesn't exist.

If $p < 1$, then the limit does not exist, because if we approach $(0,0)$ along the ray with $\phi = \frac{\pi}{4}$, we have

$$r^{p-1}\left|\cos\phi\right|\left|\sin\phi\right|^p = \left(\frac{\sqrt{2}}{2}\right)^{p+1} \frac{1}{r^{|p-1|}}$$

which is unbounded.

Thus, the limit exists iff $p > 1$, and in that case the limit is $0$.

if 1 $\geq$ p we take $(x,y)$=$(u,u)$ then we get $\frac{xy^{p}}{x^{2}+y^{2}}$=$\frac{u^{p-1}}{2}$ (=$\frac{1}{2}$ if p=1) and we can deduce that the limite don't exist.

It is indeed $p\leq 1$ (assuming $p>0$ to begin with). Let

$$f(x,y) := \frac{|x||y|^{p}}{x^2+y^2}.$$

You can show that for $p\leq 1$ we have an explicit counterexample

$$\lim_{t\to 0} f(t,t) \neq \lim_{t \to 0 }f(t,0).$$

This means that the limit $\lim_{(x,y)\to(0,0)} f(x,y)$ does not exist.

I leave the actual computation to you.