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*}
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*}
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 $\lim \inf_{(x,y)\rightarrow\infty, x\leq y} x/y = 0$).
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.
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.
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.