Question about assuming the existence of an object. I know that for $f: \mathbb{R} \rightarrow \mathbb{R}$, $f$ is cont. if and only if for any sequence $\{x_n\}_{n \ge 1} $ that converges to $z$, we have
$$\lim_{n \rightarrow \infty} f(x_n) = f(z).$$
In a proof where I have to demonstrate that $f$ is continuous, I have to assume that there exists a sequence $\{x_n\}_{n \ge 1}$ that converges to $z$.
Question: In this case, am I also allowed to assume that
$$\lim_{n \rightarrow \infty} f(x_n)  $$
exists in my proof?
Thank you.
 A: Let me first clarify a point about starting the proof. You wrote:

"I have to assume that there exists a sequence $\{x_n\}_{n \ge 1}$ that converges to $z$".

But that is not a good wording of how to start the proof, the word "exists" should not be there. Instead you should word it like this:

"I assume that $\{x_n\}_{n \ge 1}$ is a sequence that converges to $z$."

And now you must prove:

"$\{f(x_n)\}_{n \ge 1}$ converges to $f(z)$."

In doing this you are not allowed to assume that $\lim_{n \to \infty} f(x_n)$ exists.
Instead, what you must prove is that limit exists and is equal to $f(z)$.
However, it's not like you actually have to prove two different things. Any valid proof of the statement "$\{f(x_n)\}_{n \ge 1}$ converges to $f(z)$" will allow you to conclude that the limit exists and is equal to $f(z)$.
In particular, that's exactly what you may conclude if you do a proof by applying the actual $\epsilon,N$ definition of convergence:

$\{f(x_n)\}_{n \ge 1}$ converges to $f(z)$ if and only if for all $\epsilon > 0$ there exists $N$ such that if $n \ge N$ then $|f(x_n) - f(L)| < \epsilon$.

