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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.

AccidentalFourierTransform
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Shmuser
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    Are the $x$ and $y$ going to infinity at the same rate? – imranfat Sep 26 '17 at 20:23
  • @imranfat no, arbitrarily. – Shmuser Sep 26 '17 at 20:29
  • So that means that they could increase at such a rate that the denominator is zero, in such case the limit cannot exist. More over, if $x$ and $y$ increase at some variable rate to infinity, try to think of a particle moving through space, where input $x$ and $y$ are obeying the formula. Where is this particle then going to? – imranfat Sep 26 '17 at 20:32
  • I'm not sure you really understand the limit here. For one, $x=0$ cannot be a case as by the very problem statement, $x \to \infty$. – mathematics2x2life Sep 26 '17 at 20:33
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    @imranfat The denominator is equal to $(x-y)^2+y^2$, it can not be zero as $x,y$ increase. – Mike Earnest Sep 26 '17 at 20:34
  • @MikeEarnest. Good catch, but I still wouldn't understand how this limit could exist with x and y arbitrarily going up.... – imranfat Sep 26 '17 at 20:35
  • @mathematics2x2life oh, yes, thanks. – Shmuser Sep 26 '17 at 20:38
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    WolfamAlpha seems to interpret $x$ and $y$ as complex numbers. – Dabouliplop Sep 26 '17 at 21:02
  • The limit isn't perfectly clear to me, as $\infty$ isn't a single point that's being approached. So there could be numerous things that are meant. $\lim_{\lVert (x, y) \rVert \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$ is clearer, although some might consider it an abuse of notation - at least it makes clear that what you want is the limit as the norm of $(x,y)$ grows, regardless of the "angle" of $(x,y)$. – Daniel Schepler Sep 26 '17 at 22:20

5 Answers5

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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 $0.$

zhw.
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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 $$\Big|\frac{s(t-s)^2+4s^2t-s^3}{(t-s)^2+s^2}\Big|\leq\Big|\frac{(t-s)^2}{(t-s)^2+s^2}\Big||s|+\Big|\frac{s^2}{(t-s)^2+s^2}\Big||4t-s|\leq6\delta$$

Nosrati
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    shouldn't it be $3\sin t$? – bertozzijr Sep 26 '17 at 20:46
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    If $r=3\sin t + \cos t,$ then how can $r\to \infty?$ – zhw. Sep 26 '17 at 21:02
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    It is impossible for $y= r\sin t$ and $x=r(\sin t+\cos t)$ to approach $\infty$ in such a way that $r=3\sin t+\cos t$ holds. Indeed, substiting this expression for $r$ into that for $y$, you get $$y=(3\sin t +\cos t)\sin t$$so $y$ remains bounded. – Mike Earnest Sep 26 '17 at 21:19
  • The denominator should be $r^2$ in the last displayed equation, showing that the limit along this path would be zero – user795305 Sep 26 '17 at 22:10
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    There is no reason to say $t$ should go to infinity... –  Sep 26 '17 at 22:18
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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 $$ \left|\frac{(x-y)+y}{(x-y)^2+y^2}\right|\le \sqrt{2}\cdot \frac{\sqrt{(x-y)^2+y^2}}{(x-y)^2+y^2}=\sqrt{2}\cdot\frac1{\sqrt{(x-y)^2+y^2}}\le \frac{\sqrt 2}{\sqrt{y^2}} \stackrel{y\to\infty}\to 0 $$

Mike Earnest
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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 $(x,y) \rightarrow +\infty$.

Nosrati
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Julien
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    You have problems with denominators equaling $0.$ Also you can tell that if you hold $y$ at $1,$ say, then the limit of the last expression as $x\to \infty$ is $\dfrac{1}{2\sqrt 2 -2}.$ – zhw. Sep 26 '17 at 21:22
  • No because we are assuming that $(x,y) \rightarrow +\infty$ i.e. we are assuming that $x,y \geq N$ for $N$ arbitrarily large. – Julien Sep 26 '17 at 21:54
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$$ \begin{align} (x-y)^2+y^2 &=x^2-2xy+2y^2\\ &=(2-\phi)x^2+(\phi-1)x^2-2xy+\phi y^2+(2-\phi)y^2\\ &=(2-\phi)\left(x^2+y^2\right)+\left(\sqrt{\phi-1}\,x+\sqrt{\phi}\,y\right)^2\\ &\ge(2-\phi)\left(x^2+y^2\right)\\ &=\frac1{\phi^2}\left(x^2+y^2\right) \end{align} $$ Therefore, $$ \begin{align} \left|\frac{x+2y}{(x-y)^2+y^2}\right| &\le\phi^2\frac{|x+2y|}{x^2+y^2}\\ &\le\frac{3\phi^2}{\sqrt{x^2+y^2}} \end{align} $$

robjohn
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