Finding the values of n where the function is continuous at (0,0) I have to find the values of $n$ where this limit exists or doesn't exist.
$$\lim_{(x,y)->(0,0)}\frac{|x||y|^n}{x^2 + y^2}$$
I have tried to use different paths like $y=mx$ or $y=mx^2$ and such but I don'
t know what other strategies to try?
 A: Turn it into polar form:
$$\lim_{r\to0}\frac{|\cos(\theta)||\sin^n(\theta)|}{r^{1-n}}=\lim_{r\to0}r^{n-1}|\cos(\theta)||\sin^n(\theta)|$$
So the limit exists for $n>1$.
A: You need "enough degree" in the numerator to be able to cancel the denominator. 
When $n=1+c$, with $c>0$, we have (using that $|x|\leq\sqrt{x^2+y^2}$ and similar for $y$)
$$
\frac{|x|\,|y|^n}{x^2+y^2}=|y|^c\,\frac{|x|\,|y|}{x^2+y^2}
\leq|y|^c\,\frac{x^2+y^2}{x^2+y^2}=|y|^c\to0.
$$
When $n\leq1$, consider the path $x=my^n$, $x,y>0$. Then
$$
\frac{xy^n}{x^2+y^2}=\frac{my^{2n}}{m^2y^{2n}+y^2}
=\frac{m}{m^2+y^{2-2n}}\to\frac1m.
$$
So, for $n\leq1$ the limit along different path is different, and so the limit doesn't exist. 
In summary, the limit exists for $n>1$ and does not exist for $n\leq1$. 
A: $$\lim_{(x,y)\to(0,0)}\frac{|x||y|^n}{x^2 + y^2}$$
the limit is readily solved in polar form; using $\begin{cases}x=\rho\sin{\theta}\\y=\rho\cos{\theta}\end{cases}$, one obtains:
$$\lim_{\rho\to0^+}\frac{\rho^{n+1}|\sin \theta||\cos \theta|^n}{\rho^2}=\lim_{\rho\to0^+}\rho^{n-1}|\sin \theta||\cos \theta|^n=\begin{cases}\rm{undefined}&n\le1\\0&n>1\end{cases}
$$
to have a continuos function the limit has to be defined and independent of $\theta$ hence $n>1$ must be satisfied. 
if $n=1$ the function admits no limit.
