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