Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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*}

share|cite|improve this question
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 $\lim \inf_{(x,y)\rightarrow\infty, x\leq y} x/y = 0$).

share|cite|improve this answer
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

Just adding some extra comments.

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.

share|cite|improve this answer

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: \begin{equation} \lim_{x\rightarrow\infty}\lim_{y\rightarrow\infty}\frac{2x−1}{x−1}+\frac{x}{y}=2 \end{equation} and \begin{equation} \lim_{y\rightarrow\infty}\lim_{x\rightarrow y}\frac{2x−1}{x−1}+\frac{x}{y}=3. \end{equation} The non-existence of the double limit is also implicated by the non-equality of the iterated limits.

share|cite|improve this answer
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 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.