<s>[INST] 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.
 [/INST] 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
$$
\left|\frac{(x-y)+y}{(x-y)^2+y^2}\right|\le \sqrt{2}\cdot \frac{\sqrt{(x-y)^2+y^2}}{(x-y)^2+y^2}=\sqrt{2}\cdot\frac1{\sqrt{(x-y)^2+y^2}}\le \frac{\sqrt 2}{\sqrt{y^2}} \stackrel{y\to\infty}\to 0
$$
</s>