Why don't I get the correct limit of a sequence, regardless of how I arrange it (the sequence), while following the rules for solving limits? The fact that, when solving limits of sequences ($n \in \mathbb{N}$), it strikes me as very bizarre that, even though I follow all the (elementary) rules for solving limits (of sequences), I can get different results for the same example! Consider the following:
$$
\lim_{n \to \infty} \left( \frac{3n + 1}{2n - 1} \right) = \frac{\lim_{n \to \infty}(3n + 1)}{\lim_{n \to \infty} (2n - 1)} = \frac{\lim_{n \to \infty}(3n) + \lim_{n \to \infty}(1)}{\lim_{n \to \infty} (2n) - \lim_{n \to \infty}(1)} = \frac{\infty + 1}{\infty - 1} \Longrightarrow \text{This sequence does not have a limit.}
$$
I followed all of the rules for solving limits, yet this conclusion is wrong. The correct approach is the following.
$$
\lim_{n \to \infty} \left( \frac{3n + 1}{2n - 1} \right) = \lim_{n \to \infty} \left( \frac{n(3 + \frac{1}{n})}{n(2 - \frac{1}{n})} \right) = \lim_{n \to \infty} \left( \frac{3 + \frac{1}{n}}{2 - \frac{1}{n}} \right) =\frac{\lim_{n \to \infty}(3 + \frac{1}{n})}{\lim_{n \to \infty} (2 - \frac{1}{n})} = \frac{\lim_{n \to \infty}(3) + \lim_{n \to \infty}(\frac{1}{n})}{\lim_{n \to \infty} (2) - \lim_{n \to \infty}(\frac{1}{n})} = \frac{3 + 0}{2 - 0} = \frac{3}{2}.
$$
My question is: why do I get the correct answer only if I arrange the sequence a certain way?

Your reference page here starts with this introduction (emphasis mine):

Listed here are a couple of basic limits and the standard limit laws which, when used in conjunction, can find most limits. They are listed for standard, two-sided limits, but they work for all forms of limits. However, note that if a limit is infinite, then the limit does not exist.

So everywhere below that warning, if there is a statement that a limit must exist, then it's implicit that the limit must not be $\infty$.
