- You don't understand the real number system unless you understand the concept of continuity.
There are three ways to partition the rational numbers into two non-empty sets where all the numbers of the lower set are smaller than all the numbers of the higher set.

- The lower set has a largest rational number and the upper set has no smallest rational number
- The lower set has no largest rational number while the upper set has a smallest rational number
- Neither the lower set has a largest rational number nor the upper set has smallest rational number

If you replace rational numbers with the real numbers this last case cannot happen, by definition. That is what is meant by continuity -- if you cut the real number line, you are cutting at the position of a real number. While an irrational number like $$\sqrt{2}$$ cuts the number line at a place that has no rational number, so a sequence of rational numbers can forever get closer to $$\sqrt{2}$$ without ever reaching it.

- You don't understand the concept of continuity until you understand the concept of a least upper bound and a largest lower bound of an infinite but bounded set.
The sequence 1, 1/2, 1/6, 1/24, 1/120, ..., 1/n! ... goes on forever, eventually getting smaller than any positive number you can imagine. So do the sequences defined by 1/n, 1/n², 1/2ⁿ, 1/nⁿ, etc. Because in all these cases these sequences are bounded below by any non-negative number, it follows that that their greatest lower bound is 0, even though 0 is not in the set. You can explain this by taking the intersection of all upper partitions of rational number (or real numbers) where the elements of the sequence are a (proper) subset an in that way get tightest-fitting upper partition of rational numbers and see that 0 is the largest element of the lower partition. Thus 0 is the greatest lower bound of the elements of the sequence, a fact that has nothing to do with the way we order the elements of the sequence.

Limits are a bit trickier. $$a_1 = 1= \frac{p_1}{q_1} = \frac{1}{1}, \; a_{n+1} = \frac{ a_n + 2}{a_n + 1} = \frac{p_{n+1}}{q_{n+1}} = \frac{p_n + 2 q_n}{p_n + q_n}$$ defines a sequence of rational numbers:

$$1, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \frac{41}{29}, \frac{99}{70}, \dots$$ that basically settle down into a neighborhood. But there is just one such number that no matter how far you go in the sequence all the higher terms are still in successively smaller neighborhoods of that number. That number is the limit of this sequence.

Now to make my point extra pretty, I picked a sequence where the odd members are always below that number and the even members are always above it so we can partition the elements of the sequence into a lower partition whose least upper bound equals the greatest lower bound of the upper partition and this is the limit of the sequence.

Lower partition: $$1, \frac{7}{5}, \frac{41}{29}, \frac{239}{169}, \frac{1393}{985}, \dots$$

Upper partition: $$\frac{3}{2}, \frac{17}{12}, \frac{99}{70}, \frac{577}{408}, \frac{3363}{2378}, \dots$$

So hopefully, it can be seen that the only number that can be the limit of this sequence is the positive solution to $$\frac{x}{1} = \frac{x + 2}{x + 1}$$ or $$x^2 + x = x + 2$$ or $$x^2 = 2$$, because if $$p_n = \sqrt{2} q_n + \delta$$ then $$\frac{p_{n+1}}{q_{n+1}} = \frac{( 2 + \sqrt{2}) q_n + \delta}{(1 + \sqrt{2}) q_n + \delta} = \sqrt{2} + \frac{ (1 - \sqrt{2} ) \delta }{(1 + \sqrt{2} ) q_n + \delta} \approx \sqrt{2} + \frac{1 - \sqrt{2} }{1 + \sqrt{2} } \frac{\delta}{q_n}$$ where the last approximation is justified when $$\delta$$ is small compared to $$q_n$$ and $$\frac{ (1 - \sqrt{2} ) \delta }{(1 + \sqrt{2} ) q_n + \delta}$$ is therefore a smaller (in magnitude) number than $$\delta$$.