Max Payne 3 Winrar Password 💪🏿

What’s the connection between the equality of special limits and limits of sequences?

I was recently reminded of a theorem in analysis which says that if $f(x)\rightarrow_{x\rightarrow\infty}f(\infty)$, then there exists a sequence $x_n\rightarrow \infty$, such that $f(x_n)\rightarrow f(\infty)$. My question is, what is the connection between this theorem and the following one, saying that if $f(x)\rightarrow_{x\rightarrow\infty}f(\infty)$, then the sequence $(f(x_n))_{n\geq 1}$ converges to $f(\infty)$. Would it be true to write that the first theorem implies the second one?

A:

You can’t write the first theorem as a consequence of the second one (unless we assume the validity of the first, or at least that we are able to build $x_n$’s in some relevant way).
For example, the situation you describe arises when $f$ is the constant $f(x)=f(\infty)$, which is not identically constant.

A:

You don’t have to assume the first limit implies the second to have the second limit follow from the first. Also, if the two limits exist then the first limit implies the second limit, even if they exist only at certain points.
So, you can’t assume the first limit implies the second for an arbitrary function, or even any function that is only defined at some point. However, you can assume the first limit implies the second for any function in particular classes.
I guess I’ll define classes here. If $f$ is continuous, and both $f(x)$ and $f(\infty)$ exist for all $x$, then $f(x)$ converges to $f(\infty)$ at every point. If $f$ is bounded above, then $f(x)$ converges to $f(\infty)$ iff $\lim_{x \to \infty} f(x)$ exists.
That said, no matter what the limit is, there is a sequence

0644bf28c6