What are the rules for deciding which limit is quicker?

Question

For example, taking the result: $$ \lim_{M\to \infty} \lim_{J \to \infty}(1 − (1 − 2^{−M})^J) = 1 $$ I find this intuitive because for every M the inner term is smaller than 1 and thus with J→∞ we get the result, but how can this be formalized?

Answer 1

If one limit is inside the other (like here), then there is no contest between the two. You do the inner one first, see what you get, then do the outer one on that result. In this case, as long as $M$ is positive (which is the case for all $M$ we care about), we do indeed get $$ \lim_{J\to\infty}\left(1-\left(1-2^{-M}\right)^J\right) = 1 $$ so when the time comes to evaluate the $M$-limit, we get $$ \lim_{M\to \infty}1 = 1 $$

If both limits are on the "same level", things are often more interesting, and you will have to come up with some way of comparing the effects of one limit with the effects of the other, and see which one is "stronger". In this case, neither is stronger than the other to a significant enough degree that $$ \lim_{M\to \infty, J\to\infty}\left(1-\left(1-2^{-M}\right)^J\right) $$ exists. The rules for this is the $\varepsilon$-$\delta$ definition for limits (or in this case, since this time the limit is as $M, J$ goes toward $\infty$, the $\varepsilon$-$N$ definition): This last limit is equal to $1$ iff

For any $\varepsilon>0$, there is an $N\in \Bbb R$ such that whenever $N<M$ and $N<J$, we have $$ \left|\left(1-\left(1-2^{-M}\right)^J\right) - 1\right| < \varepsilon $$

And for this particular expression, that's not true because if we set, say, $\varepsilon = 0.1$, then for any $N\in \Bbb R$ and any $J>N$, you can find an $M>N$ that $\left|\left(1-\left(1-2^{-M}\right)^J\right) - 1\right|> 0.1$.