So, let me cearly state the problem:
Let $(x_n)$ be a convergent sequence, with: $ x_n > 0 $, $\forall n$, n natural number, and $x_n \to a$, with $a>0$. Then $\ln{x_n} \to \ln{a}$.
Here is my idea for a proof:
Our goal is to proof that there $\forall\epsilon>0$ there is some $n_{\epsilon}$, such that $\forall n \ge n_{\epsilon} $, we have that $|\ln{x_n}-\ln{a}|<\epsilon$.
So here is what I did. First:
$|\ln{x_n}-\ln{a}| = |\ln{\frac{x_n}{a}}| $
Then, beacause $\ln{x} <x$, $\forall x>0$, it easily follows that $\ |\ln{x}| <x$, $\forall x>0$. Applying this, we have that:
$$|\ln{x_n}-\ln{a}| = |\ln{\frac{x_n}{a}}| < \frac{x_n}{a} = \frac{x_n-a+a}{a}= \frac{x_n-a}{a} +1 $$
Now, we use the basic property of the absolute value: $x_n-a \le |x_n-a|$, that gives us:
$$|\ln{x_n}-\ln{a}| < \frac{x_n-a}{a} +1 \le \frac{|x_n-a|}{a} +1 $$
Now, we use the fact that $ x_n \to a $. So, $\forall\epsilon>0$ there is some $n_{\epsilon}$, such that $\forall n \ge n_{\epsilon} $, we have that $|x_n-a|<\epsilon$. We choose an epsilon that takes the form $a(\epsilon_0-1)$. This choice is possible for any $\epsilon_0$.
Now, we have managed to obtain that:
$$|\ln{x_n}-\ln{a}| < \frac{a(\epsilon_0-1)}{a} +1 = \epsilon_0, \forall n \ge n_{\epsilon} $$
Since $\forall \epsilon_0 > 0$ we can find an $n_{\epsilon}$, such that $\forall n \ge n_{\epsilon}$, the above inequality is staisfied, our claim is proved.
So, can you please tell me if my proof si correct? I've tried to find a proof, only for the limit of sequences! This problem has been on my nerves for s while. Also, probably there is some simpler way to do it, but I couldn't find it.
Yes, it is correct.
In fact the property $$\lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n)$$ you proved for $\ln(x)$ is called sequential continuty. It is equivalent to continuity, meaning any function $f$ is continous if and only if this holds for all series $(x)_n$.
So along the way you have shown that $\ln(x)$ is continuous on $(0,\infty)$.