Evaluating the limits $\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$ and $\lim_{(x,y)\to(\infty,8)}(1+\frac{1}{3x})^\frac{x^2}{x+y}$ I got the following problem:
Evaluate the following limits or show that it does not exist: 
$$\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$$
and
$$\lim_{(x,y)\to(\infty,8)}\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}$$
I tried for an hour and half evaluating each of those limits but I failed and I got nothing useful to share.
Some hints will be appreciated.
Thanks.
 A: By setting:

$$x:=r\cos\theta \\ y:=r\sin\theta$$

You can turn: $$\lim_{\substack{
x\rightarrow\infty\\
y\rightarrow\infty}}
{f(x,y)}$$ into: $$\lim_{r\rightarrow\infty}{g(r,\theta)}.$$
So that:

$$ \frac{2x-y}{x^2-xy+y^2}=\frac{1}{r}\cdot\frac{2\cos\theta-\sin\theta}{(1-\cos\theta\sin\theta)}\,\overset{r\rightarrow\infty}{\longrightarrow}\,0$$

The second one is easier. By noting that:
$$
\lim_{\substack{
x\rightarrow\infty\\
y\rightarrow 8}}{\frac{y}{x}} = 0
$$
We get:

$$ \left(1+\frac{1}{3x}\right)^{\frac{x^2}{x+y}}=\left(\left(1+\frac{1}{3x}\right)^{3x}\right)^{\frac{1}{3+3\frac{y}{x}}} \longrightarrow e^{\frac{1}{3}}$$

A: $$\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$$
For $y=x$ we have the following:
$$\lim_{(x,x)\to(\infty,\infty)}\frac{2x-x}{x^2-x^2+x^2}=\lim_{x \rightarrow \infty} \frac{x}{x^2}=\lim_{x \rightarrow \infty} \frac{1}{x}=0$$
If the limit exists, it will be equal to $0$.
What is left is the reasoning, that the fraction converges actually to $0$ when $x$ and $y$ go independent from each other to $\infty$.
A: Alternative solutions:
Consider the limit on all lines passing through the origin, i.e.,
$y=ax$.  $$\lim_{(x,y)\to (\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}=\lim_{x\to\infty}\frac{x\left[2-a\right]}{x^2\left[1-a+a^2\right]}=\lim_{x\to\infty}\frac{\left[2-a\right]}{x\left[1-a+a^2\right]}=0, $$ independently of the choice  of $a$.
For second question, note that 
$$\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}=e^{\frac{x^2}{x+y}\ln\left(1+\frac{1}{3x}\right) }.$$
Once that numerator and denominator are going to the infinity we can applied the L'Hopital  rule. So 
\begin{eqnarray}
\lim_{(x,y)\to (\infty,8)}\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}&=&\lim_{x\to\infty}e^{\frac{x^2}{x+8}\ln\left(1+\frac{1}{3x}\right) } \\
&=&\lim_{x\to\infty}e^{\left[2x\ln\left(1+\frac{1}{3x}\right) +x^2\frac{1}{1+\frac{1}{3x}}\frac{-1}{3x^2}\right]}\\
&=&e^{-\frac{1}{3}}\lim_{x\to \infty}\left(1+\frac{1}{3x}\right)^{2x}
\end{eqnarray}
Make $2x=y$. So $3x=(3/2)y$ and $y\to\infty$ as $x\to\infty$. Therewith
$$
\lim_{x\to \infty}\left(1+\frac{1}{3x}\right)^{2x}=\lim_{y\to\infty}\left(1+\frac{2}{3}\frac{1}{y}\right)^{y}=e^{\frac{2}{3}},\,\,\text{(for defition).}
$$
therefore 
$$
\lim_{(x,y)\to (\infty,8)}\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}=e^{-\frac{1}{3}}e^{\frac{2}{3}}=e^{\frac{1}{3}}.
$$
