How can I prove that a sequence has a given limit? For example, let's say that I have some sequence $$\left\{c_n\right\} = \left\{\frac{n^2 + 10}{2n^2}\right\}$$ How can I prove that $\{c_n\}$ approaches $\frac{1}{2}$ as $n\rightarrow\infty$?
I'm using the Buchanan textbook, but I'm not understanding their proofs at all.

Well we want to show that for any $\epsilon>0$, there is some $N\in\mathbb N$ such that for all $n>N$ we have $|c_n-1/2|<\epsilon$ (this is the definition of a limit). In this case we are looking for a natural number $N$ such that if $n>N$ then
$$\left|\frac{n^2+10}{2n^2}-\frac{1}{2}\right|=\frac{5}{n^2}<\epsilon$$
We can make use of what's called the Archimedean property, which is that for any real number $x$ there is a natural number larger than it. To do so, note that the above equation is equivalent to $\frac{n^2}{5}>\frac{1}{\epsilon}$, or $n^2>\frac{5}{\epsilon}$. If we choose $N$ to be a natural number greater than $\frac{5}{\epsilon}$, then if $n>N$ we have $n^2>N>\frac{5}{\epsilon}$ as desired. Thus $\lim\limits_{n\to\infty} c_n = \frac12$.
To relate this to the definition of limit in Buchanan: You want to show that for any neighborhood of $1/2$, there is some $N\in\mathbb N$ such that if $n>N$ then $c_n$ is in the neighborhood. Now note that any neighborhood contains an open interval around $1/2$, which takes the form $(1/2-\epsilon,1/2+\epsilon)$. Saying that $c_n$ is in this open interval is the same as saying that $|c_n-1/2|<\epsilon$.
