When does the limit $\lim_{(x,y)\to(0,0)} \frac{x^ky^l}{x^{2p}+y^{2q}}$ exist? In this case, $k,l,p,q\geq0$ and are integers. I have attempted the substitution $u=x^p$ and $v=y^q$
$$\lim_{(x,y)\to(0,0)} \frac{x^ky^l}{x^{2p}+y^{2q}}=\lim_{(u,v)\to(0,0)} \frac{u^{k/p}v^{l/q}}{u^2+v^2}$$
Then, note that $|u|<\sqrt{|u|^2+|v|^2}$ and $|v|<\sqrt{|u|^2+|v|^2}$. So we have 
$$\bigg|\frac{u^{k/p}v^{l/q}}{u^2+v^2}\bigg|=\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}<\frac{(|u|^2+|v|^2)^{k/(2p)+l/(2q)}}{|u|^2+|v|^2}$$
Thus, the limit equals $0$ when $\frac{k}{p}+\frac{l}{q}>2$. We need to show that otherwise, the limit does not exist. 
If $\frac{k}{p}+\frac{l}{q}=2$, taking the limit along the axes yields $0$ but letting $x^p=y^q$ gives 
$$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}=\frac{|u|^{k/p+l/q}}{2|u|^2}\to\frac{1}{2}$$
so the limit does not exist. 
Similarly, if $\frac{k}{p}+\frac{l}{q}<2$, taking the limit along the axes still yields $0$, but letting $x^p=y^q$ instead gives 
$$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}>\frac{|u|^{2}}{2|u|^2}\to\frac{1}{2}$$
showing that the limit does not exist.
I'm not sure if all of the steps I've taken are correct - especially some of the inequalities as $(u,v)\to(0,0)$. Also, for the last two cases, I'm not exactly sure if the absolute values around $u$ and $v$ should be there or whether I should have let $|x|^p=|y|^q$. 
 A: Let $k,l,p,q\geq0$ be integers. First, note that if $p$ or $q$ (or both) are $0$, then the denominator approaches either $1$ or $2$. If $k=l=0$, the numerator goes to $1$, otherwise it goes to $0$. In all of these cases, the limit clearly exists.
Otherwise, let $u=x^p$ and $v=y^q$, which gives
$$\lim_{(x,y)\to(0,0)} \frac{x^ky^l}{x^{2p}+y^{2q}}=\lim_{(u,v)\to(0,0)} \frac{u^{k/p}v^{l/q}}{u^2+v^2}$$
Then, note that $|u|<\sqrt{|u|^2+|v|^2}$ and $|v|<\sqrt{|u|^2+|v|^2}$. Thus we have 
$$\bigg|\frac{u^{k/p}v^{l/q}}{u^2+v^2}\bigg|=\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}<\frac{(|u|^2+|v|^2)^{k/(2p)+l/(2q)}}{|u|^2+|v|^2}$$
We can see that if when $\frac{k}{p}+\frac{l}{q}>2$, the limit equals $0$. Now we will prove that if $\frac{k}{p}+\frac{l}{q}\leq2$ otherwise, the limit does not exist. 
In the case of $\frac{k}{p}+\frac{l}{q}=2$, taking the limit along the axes yields $0$ but letting $x^p=y^q$ gives 
$$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}=\frac{|u|^{k/p+l/q}}{2|u|^2}\to\frac{1}{2}$$
so the limit does not exist in this case.
Similarly, if $\frac{k}{p}+\frac{l}{q}<2$, taking the limit along the axes yields $\infty$ if $k=l=0$, and $0$ otherwise. However, letting $x^p=y^q$ instead with $|u|<1$ (since $u\to0$) gives 
$$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}>\frac{|u|^{2}}{2|u|^2}\to\frac{1}{2}$$
showing that the limit does not exist in this case either.
Thus, the limit exists only when $p$ or $q$ (or both) are $0$ or when $\frac{k}{p}+\frac{l}{q}>2$.
