Nice Convergence Proof

This is an interesting statement and proof I think.

Here is the statement.

Suppose that lim x->inf f(x) = L and that f(n) = a_n for every integer n. Then lim a_n = L.


Proof.
Suppose lim x-> inf f(x) = L. Let e > 0, then there is a postive integer N > 0 such that for all x > N, we have,

|f(x) - L| < e.

In particular, for all positive integers n > N, we have,

|f(n) - L| < e
|a_n - L| < e.

Done.


How is this useful?

Well suppose you have a limit, like say,

lim n->inf n/e^n.

It is clear that the answer is 0 because the exponential function grows faster than any polynomial, but what if we wanted to show it? Well one solution is to use L'Hopital's rule, but we can't really use L'hopitals with sequences, so we instead think of it as,

lim x->inf x/e^x = lim x->inf 1/e^x = 0 where we used L'hopitals.


It is important to note that the converse of our original statement is NOT TRUE.

Take a_n = sin(npi). Here lim n->inf sin(npi) = lim n->inf 0 = 0, but,
lim x->inf sin(xpi) DNE.

Therefore the moral here is, you can use l'hopitals with sequences but keep in mind everything should make sense if your n is actually a real number x. If it does make sense, feel free to be abusive and use l'hopitals, most people do it, and if people do it then it's ok right?:)

I hope this has helped someone out there on the internet!