Does it make sense to define $ \aleph_{\infty}=\lim\limits_{n\to\infty}\aleph_n $? Is its cardinality "infinitely infinite"?

I recently read a book about infinity, which introduced the basic notions of different kinds of infinity. I'm a total layman concerning this topic, and one question fascinated me:

Can we, in some sense, define: $$ \aleph_{\infty}=\lim_{n\to\infty}\aleph_n $$ Such that there exists a set whose cardinal is $\aleph_{\infty}$, i.e. whose cardinality is infinitely infinite?

There are two senses of "infinite number" in play here: ordinal and cardinal. Roughly, cardinal numbers count "how many," and ordinals count "which step in a progression." The $\aleph$-numbers are cardinals. By counting $\aleph_0$, $\aleph_1$, $\aleph_2$, etc., we can see that the subscripts are ordinals, however. Just like the $\aleph$ numbers give us our infinite cardinals, we have infinite ordinals, also. If we count $0,1,2,\ldots$, there is an infinite ordinal that comes "next" after all those natural numbers; we call it $\omega$. You can keep going, and get $\omega + 1,\omega + 2,\omega+3,\ldots,\omega+\omega$, etc. The $\aleph$ numbers keep going in this same sense: after $\aleph_0,\aleph_1,\aleph_2,\ldots$, we get $\aleph_\omega,\aleph_{\omega+1},\aleph_{\omega+2},\ldots,\aleph_{\omega+\omega}$, and on and on.

$\aleph_0$ is the cardinality of the set of finite ordinal numbers $0,1,2,3,4,\ldots$,

$\aleph_1$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_0$.

$\aleph_2$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_1$.

$\aleph_3$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_2$.

and so on.

$\aleph_\omega$ is the cardinality of the set of all ordinals of cardinality $\aleph_n$ for some $n$. This is the smallest cardinal number $\ge\aleph_n$ for every finite ordinal number $n$. $\omega$ is the smallest infinite ordinal number.