Introduction of the concept of limit
Considering a sequence, the concept that has more interest in general terms is the limit of the sequence. The formal definition of this concept can seem slightly intuitive at first sight, so first we introduce the idea behind the definition.
We consider the sequence $$a_n=\dfrac{1}{n}$$.
This sequence is decreasing and lower bounded by $$0$$. Giving values we observe how the sequence gets values closer and closer to $$0$$.
This allows us to say that the elements of the sequence can be seen as points on the real straight line approaching $$0$$ as $$n$$ is being increased. More rigorously, the sequence tends to $$0$$ when $$n$$ tends to the infinite. We say that the limit of the previous sequence is $$0$$.
On the other hand, as we have already seen earlier, $$0$$ is the largest lower bound of the sequence and it coincides with the limit.
Doing an abstraction exercise, considering any decreasing sequence with lower bound $$m$$, which is the largest possible one, we can repeat the previous process and verify how the terms of the sequence tend to $$m$$. This is to say, the limit is $$m$$.
This idea is also valid for an increasing sequence with upper bound $$M$$, the smallest possible one. Then the limit of the sequence is $$M$$.
Formal limit definition
We now proceed to define the limit of a sequence through the previous idea. As we have seen, the limit of a sequence is the point $$M$$ on the real straight line if the difference between the general term of the sequence and $$M$$ is, while having increased $$n$$, as close to $$0$$ as desired.
Formalizing this, the limit of the sequence $$\{a_n\}_{n\in\mathbb{N}} $$ is the number $$a$$ if, for any fixed natural $$m$$, we can find another natural number $$N$$, so that for any $$n > N$$, the following is satisfied $$$a_n  a < \dfrac{1}{m}$$$
The absolute value is just added to simplify the notation since it is equivalent to $$$\dfrac{1}{m} < a_na < \dfrac{1}{m}$$$
If we read this definition, it says that the difference between the points of the sequence and a is smaller than $$\dfrac{1}{m}$$. Or, in other words, the difference between the points of the sequence and the number a is as small as desired.
This definition formalizes the idea given previously: we have seen, for example, that the sequence $$a_n=\dfrac{1}{n}$$ approaches $$0$$ as much as we want.
We have observed that, in fact , according to this definition it is simple to verify that the limit of the sequence $$a_n=\dfrac{1}{n}$$ is $$0$$. We have to verify how, for any $$m$$, we can find a $$N$$ so that $$\Big\dfrac{1}{n}0\Big < \dfrac{1}{m}$$ for all $$n > N$$. Choosing $$N=m$$ the property is fulfilled and $$0$$ is the limit of the sequence.
This is the formal definition corresponding to the intuitive definition of limit. And it allows us to define the limit of any sequence, even for those which are not monotonous.
For the limit definition itself, we also establish that any convergent sequence is bounded, both upper and lower, since taking $$m=1$$ we obtain $$$1+a < a_n < 1+a$$$ which is what we wanted to see.
To say that the limit of the sequence $$a_n$$ is $$a$$ we will write $$$\lim_{n \to \infty}{a_n}=a$$$
Sequences without limit and classification
Not every sequence has a limit, for example $$a_n=(1)^n$$. Intuitively, it is clear since the sequence puts between a number $$1$$ and $$1$$ and cannot approach any number, but it is necessary to formally see this fact.
Just need to choose $$m=1$$ according to the previous definition. As the sequence has two possible values only, the previous condition of limit is translated into $$1 < 1a < 1$$ and $$1 < 1a < 1$$.
Subtracting $$1$$ to the first inequality and adding $$1$$ to the second one we obtain $$2 < a < 0$$ and $$0 < a < 2$$. And, therefore, there is no $$a$$ that satisfies both conditions.
Another example of sequence that does not have a limit is the sequence $$a_n=n$$. The values of this sequence increase by $$1$$ for having increased $$n$$ plus $$1$$ so they cannot approach any number.
In contrast to the previous example, this sequence does not admit any upper bound. Following the concept of proximity of the limit we will say that the previous sequence tends to infinity. Actually, we will say that the limit is $$+\infty $$ if the sequence is not upper bounded and $$\infty$$ if it is not lower bounded.
These three examples allow us to classify the sequences in the following way;
 If the sequence has a limit we will say that it is convergent.
 If the sequence tends to infinity, as in the concept presented above, we will say that it is divergent.
 Otherwise, we will simply say that the sequence does not have a limit.
Calculation of the limit
Considering a sequence, the calculation of the limit can represent a difficult problem to be solved. We see some cases where we can calculate the limit easily.
Let's consider a sequence where the general term of the sequence is given by the quotient of two polynomials. To calculate the limit of the sequence it is enough to calculate the grade of the polynomials. Then the limit is the following one;

If the grade of the polynomial of the numerator is smaller than the grade of the polynomial of the denominator then the sequence is convergent with limit $$0$$. The example type is $$a_n=\dfrac{1}{n}$$.

If the grade of the polynomial of the numerator is greater than the grade of the polynomial of the denominator then the sequence is divergent. It tends to $$+\infty$$ or to $$\infty$$ depending on the sign of the quotient of the coefficients of the greatest grade of two polynomials. The typical examples are $$a_n=n$$ and $$a_n=n$$ with limit $$+\infty$$ and $$\infty$$ respectively.
 If the grade of the polynomial of the numerator is equal to the grade of the polynomial of the denominator then the sequence is convergent with a limit equal to the quotient of the coefficients of the largest grade of the two polynomials. The typical examples are the constant sequences, but let's see a more interesting example.
We consider the sequence $$a_n=\dfrac{3n^27n+34}{2n^2+11n+21}$$.
As the numerator and the denominator have the same grade we calculate the quotient of the coefficients of the largest grade of the two polynomials. The coefficient of the largest grade of the numerator is $$3$$ and the one in the denominator is $$2$$. Therefore the limit of the sequence is $$\dfrac{3}{2}$$.
Another case where the calculation of the limit is simple is for geometric progressions. Considering the sequence $$a_n=b^n$$ we can calculate the limit of the sequence depending on;
 If $$b > 1$$ the limit of the sequence is $$+\infty$$.
 If $$b=1$$ the sequence is constant and has limit $$1$$.
 If $$1 < b < 1$$ the limit of the sequence is $$0$$.
 If $$b\leq 1$$ the sequence has no limit.
For example, $$a_n=\Big(\dfrac{1}{2}\Big)^n$$ has limit $$0$$ since $$1 < \dfrac{1}{2} < 1$$.