Is this limit correct: $\lim_{n\to\infty} {N(n)}/{2^n}=1$?

Question

Is there a mathematical difference $n\to\infty$ between $n=\infty$ if we deduce that all infinite sequences consisting of elements $\left\{1,2\right\}$, which is for $n\in\mathbb{N}$ we have $2^n$ possible finite sequences. But, if we include $n\to\infty$ or $(n=\infty)$ we have uncountable infinite sequences.

Here, the question may seem ridiculous. I'm trying to say:

Let, $N(n)$ be a counting function of all possible sequences, for the given $n\in\mathbb{N}$.

A) Therefore, is this limit correct ?

$$\lim_{n\to\infty \\ n\in\mathbb{N}} \frac{N(n)}{2^n}=1$$

B) For $n\to\infty$, can we say $N(n)=\text{uncountable infinity}?$

Answer 1

The root of your confusion (especially in your comments to J. W. Tanner's answer) is that you're failing to distinguish between a limit and a value at infinity. You have $$ \lim_{n\to\infty} \bigl(\text{number of functions }A\to\{1,2\}\text{ when }|A|=n \bigr) = \aleph_0 $$ taken as a limit of cardinal numbers when $n$ runs through $\mathbb N$. On the other hand $$ \bigl(\text{number of functions }A\to\{1,2\}\text{ when }|A|=\aleph_0 \bigr) \ne \aleph_0 $$ In other words you can't assume that you can find a property of an infinity such as $\aleph_0$ by taking a limit of smaller cardinalities that tend to $\aleph_0$. This is not particularly mysterious; is just means that the "number of functions to $\{1,2\}$" function is not continuous at $\aleph_0$.

We can see the same effect for limits at finite real numbers; for example $$ \lim_{x\to 0} \lfloor 1-x^2\rfloor = 0 \qquad \text{but} \qquad \lfloor 1-0^0\rfloor = 1 $$

(where $\lfloor x\rfloor$ is the "largest integer $\le x$" function).

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It is often but not always the case when we have a particular function $f$ that it makes sense to define the value of $f$ "at infinity" as $\lim_{n\to\infty}f(n)$. Then the equality would hold by definition. But this is not an universal truth, just a convention of phrasing that's sometimes useful. And when our conception of $f$ already has a good meaning when we plug in an infinity, then there's no guarantee that this meaning will be the same as a limit.

Answer 2

If $N(n)=2^n$ for all $n\in\Bbb N$, then, yes, $\lim_{n\to\infty}\dfrac {N(n)}{2^n}$ is defined and equals $1$.

But remember that $\infty$ is not an element of $\mathbb N$, and you have defined $N(n)$ only for $n\in\Bbb N$.