Is $\lim_{n \to \infty} \lim_{l \to \infty} \ a_{n,l} = \lim_{l \to \infty} \lim_{n \to \infty} \ a_{n,l}\;$? Is $\lim_{n \to \infty} \lim_{l \to \infty} \ a_{n,l} = \lim_{l \to \infty} \lim_{n \to \infty} \ a_{n,l}\;$?
What happens if I replace limits with lim sups?
Thanks!
 A: Brian already answered the first question very well, so I will answer the second.  If $\lim_{x\to\infty} x_{n}$ exists, then $$\limsup{x_{n}}=\liminf{x_{n}}=\lim_{x\to\infty}x_{n}.$$  Consider what would happen if this were not the case.  Then there would exist at least two different subsequential limits.   However, since every converging sequence has each subsequential limit converge to the same limit, clearly $\lim_{x\to\infty}x_{n}$ would not exist. 
So, if the limits do, in fact, exist, then the double $\limsup$s would be equivalent to the double limits.
A: For positive integers $m$ and $n$, let $(a_{m,n})$ be the sequence with $m$ $0$'s followed by all $1$'s. Then $\lim_{m\to\infty}\lim_{n\to\infty}=1$ but $\lim_{n\to\infty}\lim_{m\to\infty}=0$.
A: $$\begin{array}{cc}
1&2&3&4&\ldots&\to&\infty\\
0&1&2&3&\ldots&\to&\infty\\
0&0&1&2&\ldots&\to&\infty\\
0&0&0&1&\ldots&\to&\infty\\
\vdots&\vdots&\vdots&\vdots&&&\vdots\\
\downarrow&\downarrow&\downarrow&\downarrow&&&\downarrow&\\
0&0&0&0&\ldots&\to&\text{OOPS!}
\end{array}$$
This PDF has an extensive discussion of the double limit, the two iterated limits, and under what conditions they’re equal.
A: The classical example to see what happens is to take 
$$
a_{m,n}=\frac{m}{m+n}.
$$
Then
$$
\lim_m\lim_n a_{m,n}=0,\ \ \lim_n\lim_n a_{m,n}=1.
$$
For the sup we have, for any double sequence $b_{m,n}$, 
$$
\sup_m\sup_n b_{m,n}=\sup_{m,n}b_{m,n}=\sup_n\sup_m b_{m,n}.
$$
