Limit Definition for Proof How can I use the definition of a limit to set up a proof for a statement such as:
$$\lim_{n\to\infty} {f(n)} \to \infty$$
I have tried applying the standard definition, but I come out with meaningless expressions.
 A: To deal with infinite limits being infinite, go back to the $\epsilon$-$\delta$ definition of a limit, and adjust the parameters. What do you want to say in principle? You want to say that as my parameter becomes arbitrarily large, my value becomes arbitrarily large. The statment for a function $\lim_{x\rightarrow\infty}f(x)=\infty$ should look like this:
Given $M>0$ (large), I can find an $N>0$ such that $x>N \Rightarrow f(x)>M$.
A: Definition: Let $f$ be a function with domain $D\subseteq\mathbb{R}$, which contains arbitrarily large values. This function $f$ has a limit as $x$ approaches (tends to) plus infinity, that is to $\infty$, also denoted $+\infty$, if and only if there exists a real number $L$ such that for every $\epsilon>0$, there exists a real number $M>0$ such that
$$
|f(x)-L|<\epsilon\qquad\text{if $x\geq M$ and $x\in D$}.
$$
If such an $L$ exists, we say that $L$ is the limit of the function $f$ as $x$ tends to $\infty$, or simply, $L$ is the limit of $f$ at plus infinity, or $L$ is the limiting value at $\infty$, and we write
$$
\lim_{x\to\infty} f(x)=L,\tag{1}
$$
where $x$ is a dummy variable. 
Example: Assume $D=\mathbb{R}\setminus\mathbb{Q}$ and let $f:D\to\mathbb{R}$ be a function defined by $f(x)=\frac{x}{x+2}$. Verify that $\lim_{x\to\infty}\frac{x}{x+2}=1$.
Solution. We begin with an arbitrary but fixed $\epsilon >0$, and then verify that $(1)$ is true with $L=1$. We need to find a positive real number $M$ so that $|f(x)-1|<\epsilon$ whenever $x\geq M$, provided that $x\in D=\mathbb{R}\setminus\mathbb{Q}$. To find this $M$, we write
$$
|f(x)-1|=\left| \frac{x}{x+2}-1\right|=\frac{2}{x+2}<\frac{2}{x}
$$
if $x>0$, say. Since we want $|f(x)-1|<\epsilon$, consider $\frac{2}{x}<\epsilon$ and solve for $x$. Hence, $x>\frac{2}{\epsilon}$ and $x>0$. Hence, pick any $M>\frac{2}{\epsilon}$, which is automatically positive. Then, for all $x\geq M$ with $x\in D$, we have
$$
|f(x)-1|=\frac{2}{x+2}<\frac{2}{x}\leq\frac{2}{M}<\epsilon.
$$
Thus, we have verified the value of $\lim_{x\to\infty}f(x)$ using the definition given in $(1)$. 

In the specific example I gave, observe that if $D$ were bounded, then $\lim_{x\to\infty}f(x)$ would not exist. 
