Concept behind the limit to infinity? I can across transfinite numbers and came up with a thought.
What if$$\lim_{x\to\infty}f(x)=f(T)$$where $T$ was a transfinite number?
Generally, in calculus, I have noted that it is two different things, $\lim_{x\to a}f(x)\ne f(a)$.  Which makes me wonder about the situation $a=\infty$.  Which means it must mean something when we compare $\lim_{x\to\infty}f(x)$ and $f(\infty)$.  The two being different, I guessed the above statement about limits to infinite and transfinite numbers.
Because by definition:$$R<T<\infty$$where $R$ is the real numbers.
By definition:$$\lim_{x\to\infty_1^-}x<\infty_1$$Where the subscript indicates the two infinities are equal.  In general, the limit from the left side is close, but always less than the actual value.  At least, that is my understanding.
But at the same time, a limit to infinite is a limit by which $x$ exceeds real numbers.
So in a sense, $\lim_{x\to\infty}f(x)=f(T)$.
My question is whether or not my "postulation" is correct.
I have also noted that sometimes, on very rare occasions, the following is true:$$\lim_{x\to\infty}f(x)\ne f(\infty)$$
I will leave this up to you guys.

Here is (at least) one problem with this conception.  In the case of something as simple as, say,
$$
L = \lim_{x \to \infty} \frac{x^2+1}{2x^2}
$$
we should be able to use the conception to determine that $L = 1/2$.  And yet, if we plug in any transfinite number for $x$, we find ourselves unable to perform the arithmetic in any useful way to obtain $L = 1/2$.  Any transfinite quantity $T$ we plug in yields the indeterminate form
$$
L = \frac{T}{T}
$$
It doesn't matter so much whether it's "correct" in some Platonic sense as much as whether we can get any use out of it.  Without more of a framework behind your conception, it sure seems at first flush that the answer is no, we can't.
