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

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)?

• In the example you gave, neither limit exists. But you might consider $f(x) = \sin(\pi x)$ and $s_n = \sin(\pi n)$. Jun 8 at 20:29

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.

• Thanks for your response! I wasn't aware that the contrapositive to the statement "If $\lim_{x\to\infty} f(x) = L$, then $\lim_{n\to\infty} s_n = L$." is true, so that's good to know. It took me some time to understand why your example is true since I had to think about how $x \in \mathbb{R}$, but $n \in \mathbb{N}$. Jun 8 at 21:00
• @Anonymous73648 Note that the contrapositive of a true statement is always true.
– Stef
Jun 9 at 11:06

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 $$L$$.