When can we interchange limits of functions with limits of sequences?

Question

I just noticed how the definitions for $\lim\limits_{x\to\infty}f(x) = L$, and $\lim\limits_{n\to\infty}s_n = s$ are essentially the same thing.

Because of this, are we allowed to "swap out" $\lim\limits_{x\to\infty}f(x)$ with $\lim\limits_{n\to\infty}s_n$ if {$s_n$} is the equivalent sequence of $f(x)$, regardless of whether or not the limit exists? For instance, if we wanted to evaluate $\lim\limits_{x\to\infty}$sin(x), is it okay to instead consider $\lim\limits_{n\to\infty}s_n$, where $s_n$ = sin(n)?

Answer 1

Restricting a function of a real variable to its values on a discrete set can lose information. So we can't swap one for the other willy-nilly.

By “equivalent sequence” I guess you mean setting $s_n = f(n)$. Here are some things you can say:

• If $\lim_{x\to\infty} f(x) = L$, then $\lim_{n\to\infty} s_n = L$.
• Contrapositively, if $\lim_{n\to\infty} s_n$ does not exist, then $\lim_{x\to\infty} f(x)$ does not exist.

But now consider the function $f(x) = \sin \pi x$. Setting $s_n = f(n)$, we see that $\lim_{n\to\infty} s_n = \lim_{n\to\infty} 0 = 0$. But $\lim_{x\to\infty} f(x)$ does not exist. So it is possible for the sequence to have a limit while the function does not.

Answer 2

You’re on the right track. We can say a limit exists at a point in the extended real numbers iff for every sequence $\{x_n\}$ converging to that point, then $\{f(x_n)\}$ converges to $L$.