Is the limit $\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$ zero? I have this limit:$$\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$$
At first sight seems that limit equals 0. But WolframAlpha says that there is no limit. I tried to prove it. I considered cases $y = kx$, and so on. I never got to find subsequence, whose has limit $\neq 0$. 
I think there is a problem in denominator. When $x\rightarrow \infty$ and $y \rightarrow \infty$ we got there $\infty - \infty + \infty$. It's unclear what to do with it and how to find necessary subsequence.
Maybe i'm on the wrong way to solve it. Please, give me a tip.
 A: The expression equals
$$\frac{x+2y}{(x-y)^2 + y^2}.$$
Go to the usual polar coordinates: $x=r\cos t,y=r\sin t.$ The expression then equals
$$\tag 1 \frac{1}{r}\frac{\cos t + 2\sin t}{(\cos t-\sin t)^2 + \sin^2t}.$$
Now the denominator is nonnegative and never $0,$ and therefore is bounded below by a positive constant $c.$ It follows that the absolute value of $(1)$ is bounded above by
$$\frac{1}{r}\cdot \frac{3}{c}.$$
As $(x,y)\to \infty$ (whatever that means, I'm guessing it means $(x^2+y^2)^{1/2} \to \infty$), we must have $r\to \infty,$ so the limit is $0.$
A: For
$$\lim_{\substack{x\rightarrow\infty \\y\rightarrow\infty}} \frac{x+2y}{(x - y)^2 + y^2}$$
Let $x-y=r\cos t$ and $y=r\sin t$ so the limit
$$\lim_{\substack{r\rightarrow\infty \\t\rightarrow\infty}}\dfrac{3\sin t+\cos t}{r}$$
doesn't exist, since you can move along $r=3\sin t+\cos t$.
Update:
As people said, this approach is wrong actually, since $|r|\leq\sqrt{10}$ won't go to $\infty$, but I think with $x=\dfrac1s$ and $y=\dfrac1t$ then
$$\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}=\lim_{(s,t) \to (0,0)} \frac{st^2+2s^2t}{(t-s)^2+s^2}=\lim_{(s,t) \to (0,0)} \frac{s(t-s)^2+4s^2t-s^3}{(t-s)^2+s^2}$$
so the limit will be $0$ since
$$\Big|\frac{s(t-s)^2+4s^2t-s^3}{(t-s)^2+s^2}\Big|\leq\Big|\frac{(t-s)^2}{(t-s)^2+s^2}\Big||s|+\Big|\frac{s^2}{(t-s)^2+s^2}\Big||4t-s|\leq6\delta$$
A: It seems to me like the limit is zero. Write the function as
$$
\frac{(x-y)+y}{(x-y)^2+y^2}+\frac{2y}{(x-y)^2+y^2}
$$
The second fraction goes to zero since
$$
\left|\frac{2y}{(x-y)^2+y^2}\right|=\frac{|2y|}{(x-y)^2+y^2}\le \frac{2|y|}{y^2}=\frac2{|y|}\stackrel{y\to\infty}\to 0
$$
For the first fraction, use the Cauchy-Schwarz inequality:
$$
|(x-y)\cdot1+y\cdot1|\le \sqrt{(x-y)^2+y^2}\cdot \sqrt{1^2+1^2}
$$
Then
$$
\left|\frac{(x-y)+y}{(x-y)^2+y^2}\right|\le \sqrt{2}\cdot \frac{\sqrt{(x-y)^2+y^2}}{(x-y)^2+y^2}=\sqrt{2}\cdot\frac1{\sqrt{(x-y)^2+y^2}}\le \frac{\sqrt 2}{\sqrt{y^2}} \stackrel{y\to\infty}\to 0
$$
A: An easier answer maybe:
$$\frac{x+2y}{x^{2}-2xy+2y^{2}} = \frac{x+2y}{(x^{2}+2y^{2})-2xy} \leq \frac{x+2y}{2\sqrt{2}xy -2xy} = \frac{1}{(2\sqrt{2}-2)y} + \frac{1}{(\sqrt{2}-1)x}$$ by noting that the denominator is positive (equal to $(x-y)^{2}+y^{2}$) and by using $x^{2}+2y^{2} \geq 2\sqrt{2}xy$.
Now this obviously tends toward $0$ as $(x,y) \rightarrow +\infty$.
A: $$
\begin{align}
(x-y)^2+y^2
&=x^2-2xy+2y^2\\
&=(2-\phi)x^2+(\phi-1)x^2-2xy+\phi y^2+(2-\phi)y^2\\
&=(2-\phi)\left(x^2+y^2\right)+\left(\sqrt{\phi-1}\,x+\sqrt{\phi}\,y\right)^2\\
&\ge(2-\phi)\left(x^2+y^2\right)\\
&=\frac1{\phi^2}\left(x^2+y^2\right)
\end{align}
$$
Therefore,
$$
\begin{align}
\left|\frac{x+2y}{(x-y)^2+y^2}\right|
&\le\phi^2\frac{|x+2y|}{x^2+y^2}\\
&\le\frac{3\phi^2}{\sqrt{x^2+y^2}}
\end{align}
$$
