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?
 A: 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 $\infty$.
A: You did not in fact follow all the rules (and this is a good example of why understanding why the rules are true is important). The relevant rule is:

If both $\lim_{n\rightarrow a}f(n)$ and $\lim_{n\rightarrow a}g(n)$ exist and are finite (and $\lim_{n\rightarrow a}g(n)\not=0$), then $$\lim_{n\rightarrow a}{f(n)\over g(n)}={\lim_{n\rightarrow a}f(n)\over \lim_{n\rightarrow a}g(n)}.$$

There are various other rules of similar flavor. But the point is that the "rule" you've tried to apply is not one of them. (And it's a good exercise at this point to go through your textbook and look at what the rules do in fact say, and note that none of them actually get you what you want.)
Indeed, the example you give is a good example of the limitations of these rules: while we can often manipulate limits "algebraically," we can't do this in all cases, and some hypotheses are needed. It's also a good example of why proofs are important, since there are plenty of plausible-sounding statements which are in fact false.
