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.
 A: You need to show that for any number $\epsilon>0$ you can find $N$ such that if $n\geq N$, then $|c_n - \frac{1}{2}| < \epsilon$.
So, pick an $\epsilon>0$ and start computing: $|c_n - \frac{1}{2}| = | \frac{n^2+10}{2 n^2} - \frac{1}{2}| = |\frac{10}{2 n^2}| = \frac{5}{n^2}$.
Now you need to pick $N$ such that if $n \geq N$, then $\frac{5}{n^2} < \epsilon$. If I pick $N \geq \sqrt{\frac{5}{\epsilon}}+1$, then you can see that if $n \geq N$, then $\frac{5}{n^2} < \epsilon$. Since $\epsilon>0$ was arbitrary, you are finished.
A: 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 $|c_n-1/2|<\epsilon$.
A: $$
\lim_{n \rightarrow \infty} \frac {n^2+10}{2n^2} = \lim_{n \rightarrow\infty} \frac {1+\frac {10}{n^2}}{2} = \lim_{n \rightarrow \infty} \left (\frac 12+\frac{10}{2n^2} \right) = \frac 12+5\lim_{n \rightarrow \infty}\frac 1{n^2}=\frac 12
$$
A: In general, we say that a sequence $x_n \rightarrow x$ if $|x_n - x| \rightarrow 0$. This means that for all $\epsilon > 0$ there exists some $N \in \mathbb{N}$ such that for all $n \geq N$, $|x_n - x| < \epsilon$.
We can also use results about sequences that will be proved in any good introductory texts, such as if $x_n \rightarrow x$ and $y_n \rightarrow y$, $\lambda \in \mathbb{R}$: 
$i) x_n + y_n \rightarrow x+y$
$ii) x_ny_n \rightarrow xy$
$iii)\lambda x_n \rightarrow \lambda x$
$iv)$ If $x \not= 0$ then  $\frac{y_n}{x_n} \rightarrow \frac{y}{x}$
Using these results we can see that $\displaystyle c_n = \frac{n^2 + 10}{2n^2} = \frac{1+\frac{10}{n^2}}{2}$. So if $a_n = \frac{10}{n^2} \rightarrow a$ then $c_n \rightarrow \frac{1+a}{2}$. Now we can prove that $a_n \rightarrow 0$ using epsilon delta techniques, so it follows that $c_n \rightarrow \frac12$.
