# What is a conditional double limit and how to compute it?

What is a conditional double limit and how to compute it for

\begin{equation*} \lim_{(x,y)\rightarrow\infty,x\leq y}\frac{2x-1}{x-1}+\frac{x}{y}? \end{equation*}

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What's $k$ supposed to be? – J. M. Sep 12 '10 at 23:19
Actually, $k$ should be $x$. I corrected that. – Oleksandr Bondarenko Sep 13 '10 at 0:22
You should state your definition for $\lim_{(x,y)\rightarrow \infty} f(x,y)$, because it's not natural. I mean, on a plain, what it $\infty$? Very far on right top-righthand corner, top-lefthand? We usually defined the limit to infinity as $\left\| (x,y) \right\| \rightarrow \infty$. – Andy Sep 13 '10 at 10:08

The limit depends on the exact way $x,y$ approach infinity. The first summand $(2x-1)/(x-1)$ actually tends to $2$ (always). The second summand can tend to any limit in $[0,1]$, or to no limit at all (in which case the limit of the entire sum does not exist). To get a limit of $0$, use $y = x^2$. To get a limit of $0 < a\leq 1$, use $x = ay$. To get no limit, you can use $x = y(sin(y)+2)/3$ [thanks AD!], as well as many other options.

I suppose you could define $\lim \sup$ as the supremum of all limits attainable for some sequence $(x_n,y_n)$, and then $\lim \sup_{(x,y)\rightarrow\infty, x\leq y} x/y = 1$ (and similarly $\lim \inf_{(x,y)\rightarrow\infty, x\leq y} x/y = 0$).

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The example $x=y|\sin y|$ does not work since such $x$ does not tend to infinity - $\lim_{(x,y)\to\infty, x\le y}$ means that both $x$ and $y$ should tend to $\infty$ and that $x\leq y$. You can use $x= y^{(\sin(y)+2)/3}$ or something similar to get no limit. – AD. Sep 13 '10 at 5:32
I can not reedit..exponent is not necessary. One can use $x= y(\sin y + 2)/3$. Easier! – AD. Sep 13 '10 at 5:41
If I correctly understood in the result we get the following: $\lim\sup_{(x,y)\rightarrow\infty,x\leq y}\frac{2x-1}{x-1}+\frac{x}{y}=3$ and $\lim\inf_{(x,y)\rightarrow\infty,x\leq y}\frac{2x-1}{x-1}+\frac{x}{y}=2$? – Oleksandr Bondarenko Sep 13 '10 at 9:14

First $$\frac{2x-1}{x-1}=\frac{2-1/x}{1-1/x}\to\frac{2-0}{1-0}=2$$ as $x\to\infty$ (this is independent of how the limit is taken). Secondly, we note that $$0\leq\frac{x}{y}\leq1$$ the first inequality comes from that both $x,y$ are $>0$ (since they should tend to $\infty$), the second comes from $x\leq y$ (divide by $y$).

At this point we have $$2\leq\liminf_{(x,y)\to\infty,x\leq y} \frac{2x-1}{x-1} + \frac{x}{y}\leq\limsup_{(x,y)\to\infty,x\leq y} \frac{2x-1}{x-1} + \frac{x}{y}\leq3$$ At the boundary of the limit, that is when $x=y$ we would get 3. To reach the liminf we only have to find a way to press $x/y$ as low as we can (recall that we already know it is $x/y\ge0$) - at this point you have do some testing, you want $x$ to tend slower to infinity than $y$, $x=\log y$ or $x=\sqrt{y}$ works fine!

Also, the linear choice $x=ay$, earlier suggested by Yuval Filmus above works fine to see that all limits between 2 and 3 are possible. The example in the the comment of Yuval Filmus shows that there are choices of $x$'s such that no limit is attained.

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It seems to me that the limit of the function does not exist, since the function approaches to different limits along different ways. However, we can compute iterated limits: $$\lim_{x\rightarrow\infty}\lim_{y\rightarrow\infty}\frac{2x−1}{x−1}+\frac{x}{y}=2$$ and $$\lim_{y\rightarrow\infty}\lim_{x\rightarrow y}\frac{2x−1}{x−1}+\frac{x}{y}=3.$$ The non-existence of the double limit is also implicated by the non-equality of the iterated limits.

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Sure you can, just remember to stay in the track - $x\le y$ that is (you should have $x\to y$ and then let $y\to\infty$ in your second limit). – AD. Sep 13 '10 at 20:25
BTW.. Welcome to math.stackexchange.com where we all teach and learn, I hope. =) – AD. Sep 13 '10 at 20:26
@AD.:Concerning the second limit, of course, you are right. So I corrected $x\rightarrow\infty$ to $x\rightarrow y$. What I obtained is that the second iterated limit tends to three. Am I right? – Oleksandr Bondarenko Sep 13 '10 at 22:21
Exactly! By luck we got the $\liminf$ and the $\limsup$. – AD. Sep 14 '10 at 5:25

Leave the first term (with limit $2$) on the side.

Then by letting $x=py$ with $p\le1$, you can achieve any $p$, so the limit does not exist.

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