For what $m,n$ does the limit exist. Let $f: (0, \infty) \times (0, \infty) \to [0, \infty)$ by given by $$f(x,y) = \frac{x^ny^m}{x+y}.$$ Find all $m,n$ such that $\lim_{(x,y) \to (0,0)} f(x,y)$ exists. 
Consider the line $y=kx$ for some $k \in \mathbb{R}$. We observe that \begin{eqnarray*}
\lim_{(x,y) \to (0,0)} \frac{x^ny^m}{x+y} &=& \lim_{(x,kx) \to (0,0)} \frac{x^n(kx)^m}{x+(kx)} \\
&=& \lim_{(x,kx) \to (0,0)} \frac{k^m x^{n+m}}{x(1+k)} \\
&=& \lim_{(x,kx) \to (0,0)} \frac{k^m}{1+k} \cdot x^{n+m-1}.
\end{eqnarray*} This diverges if $n+m-1 < 0$. Also, the limit is dependent on $k$ if $n+m-1=0$. Therefore, we see that $\lim_{(x,y) \to (0,0)} f(x,y)$ exists if $n+m -1 > 0$. 
My question is regarding sufficiency, sure, my solution is necessary, but it is sufficient to simply take the path $y=kx$?
 A: No it is not sufficient. There are plenty of functions $f(x,y)$ defined in that region that have the same limit along every line $y = kx$, but whose limit does not exist, for example, $f(x,y) = \frac{xy^2}{x^2 + y^4}$. You need to show it rigorously, which for this function is not that difficult.  
A: Transforming to polar coordinates $(\rho,\phi)$, we have
$$\begin{align}
\frac{x^ny^m}{x+y}&=\rho^{n+m-1}\frac{\cos^n(\phi)\sin^m(\phi)}{\sqrt{2}\sin(\phi+\pi/4)}
\end{align}$$
Now, note that for $x\ge 0$ and $y\ge 0$, $0\le \phi\le \pi/2$.  Therefore, we find that  
$$\left|\frac{\cos^n(\phi)\sin^m(\phi)}{\sqrt{2}\sin(\phi+\pi/4)}
\right|\le 1$$
Therefore, the limit of interest is $0$ when $n+m-1>0$, but does not exist otherwise (i.,e., the limit is path dependent).
A: You did not show that the limit exists when $m,n$ are non-negative and $m+n>1.$
For this case, let $z=\max(x,y).$ We have $(x+y)^{-1}<z^{-1}$ and $x^n y^m\leq z^{n+m}.$ So $$0<x^n y^m (x+y)^{-1}<z^{n+m}z^{-1}=z^{m+n-1}.$$ And $\lim_{(x,y)\to (0,0)}z=0,$ so the limit of $f(x,y)$ is $0.$
You can also show the limit does not exist when $n<0$ or $m<0$. 
