Limit of a Sequence
Consider the sequence $\left \{ \frac{1}{n} \right \}_{n=1}^{\infty} = \left \{ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ... \right \}$. As $n \to \infty$, it appears as though $\frac{1}{n} \to 0$. In fact, we know that this is true since $\lim_{n \to \infty} \frac{1}{n} = 0$. We will now formalize the definition of a limit with regards to sequences.
Definition: If $\{ a_n \}$ is a sequence, then $\lim_{n \to \infty} a_n = L$ means that for every $\epsilon > 0$ there exists a corresponding $N \in \mathbb{N}$ such that if $n ≥ N$, then $\mid a_n  L \mid < \epsilon$. If this limit exists, then we say that the sequence $\{ a_n \}$ Converges, and if this limit doesn't exist then we say say the sequence $\{ a_n \}$ Diverges.

We note that our definition of the limit of a sequence is very similar to the limit of a function, in fact, we can think of a sequence as a function whose domain is the set of natural numbers $\mathbb{N}$. From this notion, we obtain the very important theorem:
Theorem 1: If $\{ a_n \}$ is a sequence and a function $f(x)$, then if $f(n) = a_n$ where $n \in \mathbb{N}$ and $\lim_{x \to \infty} f(x) = L$, then $\lim_{n \to \infty} f(n) = L$. 
 Proof of Theorem 1: Let $\epsilon > 0$ be given. We know that $\lim_{x \to \infty} f(x) = L$ which implies that $\forall \epsilon > 0 \: \exists N \in \mathbb{R}$ such that if $x ≥ k$ then $\mid f(x)  L \mid < \epsilon$.
 Now we want to show that $\forall \epsilon > 0 \: \exists N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n  L \mid < \epsilon$. We will choose $N = \mathrm{max} \{ 1, \lceil k \rceil \}$. This ensures that $N$ is an integer.
 Now since $N ≥ k$ then it follows that if $\forall n ≥ N$ then $\mid f(n)  L \mid < \epsilon$. But $f(n) = a_n$ and so $\mid a_n  L \mid < \epsilon$ so $\lim_{n \to \infty} a_n = L$. $\blacksquare$
Important Note: The converse of this theorem is not implied to be true! That is if $f(n) = a_n$ and $\lim_{n \to \infty} f(n) = L$, then this does NOT imply that $\lim_{x \to \infty} f(x) = L$. For example, consider the sequence $\{ \sin (2\pi n) \} = \{ 0, 0, 0, ... \}$. Clearly this sequence converges at 0. However, the function $f(x) = \sin (2\pi x)$ does not converge, instead, it diverges as it oscillates between $1$ and $1$. 
For example, consider the sequence $\left \{ \frac{n + 2}{n} \right \}_{n=1}^{\infty}$. If we let $f(n) = \frac{n + 2}{n}$ be a function whose domain is the natural numbers, then we calculate the limit of this function like we have in the past, namely:
(1)
\begin{align} \: \lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{n + 2}{n} = \lim_{n \to \infty} 1 + \frac{2}{n} = 1 \end{align}
Therefore the limit of our sequence $f(n) = a_n$ is 1, that is, $\left \{ \frac{n + 2}{n} \right \}_{n=1}^{\infty}$ converges to 1 as $n \to \infty$.
Now let's look at another major theorem.
Theorem 2: If $\lim_{n \to \infty} a_n = L$ and a function $f$ is continuous at $L$, then $\lim_{n \to \infty} f(a_n) = f \left(\lim_{n \to \infty} a_n \right) = f(L)$. 
 Proof of Theorem 2: If $f$ is a continuous function at $x = L$, then we know that $f(L)$ is defined and that $\lim_{x \to L} f(x) = f(L)$. By the definition of a limit, $\forall \epsilon > 0 \: \exists \delta > 0$ such that if $0 < \mid x  L \mid < \delta$ then $\mid f(x)  f(L) \mid < \epsilon$. Let $x = a_n$ so then if $0 < \mid a_n  L \mid < \delta$ then $\mid f(a_n)  f(L) \mid < \epsilon$.
 We want to show that $\lim_{n \to \infty} f(a_n) = f(L)$, that is $\forall \epsilon > 0 \: \exists N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid f(a_n)  f(L) \mid < \epsilon$, which is what we showed above.
We will now look at some important limit laws regarding sequences
Limit Laws of Convergent Sequences
We will now look at some very important limit laws regarding convergent sequences, all of which are analogous to the limit laws for functions that we already know of.
Law 1 (Addition Law of Convergent Sequences): If the limits of the sequences $\{ a_n \}$ and $\{ b_n \}$ are convergent, that is $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ then $\lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = A + B$. ( Proof of Law 1 ) 
Law 2 (Difference Law of Convergent Sequences): If the limits of the sequences $\{ a_n \}$ and $\{ b_n \}$ are convergent, that is $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ then $\lim_{n \to \infty} (a_n  b_n) = \lim_{n \to \infty} a_n  \lim_{n \to \infty} b_n = A  B$. ( Proof of Law 2 ) 
Law 3 (Product Law of Convergent Sequences): If the limits of the sequences $\{ a_n \}$ and $\{ b_n \}$ are convergent, that is $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$, then $\lim_{n \to \infty} (a_nb_n) = \lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = AB$. ( Proof of Law 3 ) 
Law 4 (Quotient Law of Convergent Sequences): If the limits of the sequences $\{ a_n \}$ and $\{ b_n \}$ are convergent, that is $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$, then $\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{\lim_{n \to \infty} a_n}{\lim_{n \to \infty} b_n} = \frac{A}{B}$ provided that $\lim_{n \to \infty} b_n ≠ 0$. ( Proof of Law 4 ) 
Law 5 (Constant Multiple Law of Convergent Sequences): If the limit of the sequence $\{ a_n \}$ is convergent, that is $\lim_{n \to \infty} a_n = A$, and $k$ is a constant, then $\lim_{n \to \infty} ka_n = k \lim_{n \to \infty} a_n = kA$. ( Proof of Law 5 ) 
Law 6 (Power Law of Convergent Sequences): If the limit of the sequence $\{ a_n \}$ is convergent, that is $\lim_{n \to \infty} a_n = A$ and $k$ is a nonnegative integer, then $\lim_{n \to \infty} (a_n)^k = \left ( \lim_{n \to \infty} a_n \right )^k = (A)^k$ provided that $k \in \mathbb{N}$. ( Proof of Law 6 ) 
Law 7 (Squeeze Theorem for Convergent Sequences): If the limits of the sequences $\{ a_n \}$, $\{ b_n \}$, and $\{ c_n \}$ are convergent and $a_n ≤ b_n ≤ c_n$ is true always after some $n^{\mathrm{th}}$ term, if $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L$, then $\lim_{n \to \infty} b_n = L$. ( Proof of Law 7 ) 
Example 1
Determine whether the sequence $\left \{ \frac{x^2  1}{x + 1} \right \}_{n=1}^{\infty}$ is convergent or divergent.
Let $f(n) = \frac{n^2  1}{n + 1}$ be a function analogous to our sequence. When we factor the numerator and cancel liketerms, we get that $\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{n^2  1}{n + 1} = \lim_{n \to \infty} \frac{(n +1)(n  1)}{n + 1} = \lim_{n \to \infty} n  1 = \infty$. Therefore the sequence $\left \{ \frac{x^2  1}{x + 1} \right \}_{n=1}^{\infty}$ is divergent.