What is the Epsilon Delta Algebraic Statement for limits at Infinity and Infinite Limits

For the standard Limit: $\lim_{x\to c}[f(x)] = L$

This is the statement used to express limits in Algebraic form

$\forall \epsilon > 0 \hspace{3mm} \exists\delta > 0 $ such that $|x-c|< \delta \implies |f(x) - L|<\epsilon$

I'm wondering what are the standard limit statements used for both infinite limits and limits at infinity:

$$\lim_{x \to \infty } [f(x)] = L \hspace{10mm} \lim_{x \to c} [f(x)] = \infty$$

I know that letters M and N are involved but I don't know in what way. Let me know if I made any errors.

Following @podiki's comment, Wikipedia's definition for $\lim_\limits{x\to+\infty}f(x)=L$ is:

$$ \forall\epsilon>0,\exists N,\forall x>N:\left|f(x)-L\right|<\epsilon. $$

To interpret this statement (and link it to the 'standard' definition for when $x\to c$ for some $c\in\mathbb{R}$), we need to talk about neighborhoods.

For any real number $x$, a neighborhood of $x$ (for our purposes...) is simply an open interval containing $x$. For example, if $\delta$ is any positive real number, then $\left(x-\delta,x+\delta\right)$ is a neighborhood of $x$. Another way to specify a neighborhood of $x$ is to specify a condition for other real numbers to be in them. For a real number $y$ to be in $\left(x-\delta,x+\delta\right)$ is equivalent to saying that $\left|x-y\right|<\delta$. (Check that they are indeed equivalent!)

The statement $\lim_\limits{x\to c}f(x)=L$ then reads: "For every neighborhood (call it $V$) of $L$, there exists a neighborhood (call it $U$) around $c$ such that if $x\in U$, then $f(x)\in V$". Intuitively, no matter how closely we look at $L$ - say, the neighborhood $V=\left(L-\epsilon,L+\epsilon\right)$ - there will always be some small enough neighborhood $U=\left(c-\delta,c+\delta\right)$ around $c$ such that for all $x\in U$, $f(x)\in V$. This captures the intuitive notion of '$f(x)$ gets closer to $L$ as $x$ approaches $c$'.

Now, we also want to make sense of the case when $x$ gets arbitrarily large... whats a neighborhood around $\infty$? Well, $\infty$ is not a real number, so no open interval can 'contain' $\infty$. We need to generalize our notion of an open interval to include this case. Recall that an open interval $(a,b)$ in $\mathbb{R}$ is the set $\left\{x\in\mathbb{R}\;|\;a<x<b\right\}$. Define an 'open ray' or a 'neighborhood of $+\infty$' in $\mathbb{R}$ as any set $(N,+\infty):=\left\{x\in\mathbb{R}\;|\;x>N\right\}$. Extend the notion of a neighborhood around $x$ to include open rays as well.

The statement $\lim_\limits{x\to+\infty}f(x)=L$ then reads: "For every neighborhood $V=\left(L-\epsilon,L+\epsilon\right)$ of $L$, there exists a neighborhood $U=(N,+\infty)$ of $+\infty$ such that if $x\in U$, then $f(x)\in V$". See the connection?

Personally, I find the $\epsilon-\delta$ definitions for limits cumbersome. However, translating definitions between neighborhoods and $\epsilon,\delta$'s is a good skill to have.

Exercise: Try to formulate the statements $\lim_\limits{x\to c}f(x)=+\infty$ and $\lim_\limits{x\to+\infty}f(x)=+\infty$.

Note: As @Snaw commented, we need to specify that $x\neq c$ (this is implied from the statement $0<|x-c|<\delta$) for limits to some $c\in\mathbb{R}$ and $x\neq N$ (since $x>N$) for limits to infinity. These are more technical restrictions, and IMO does not hinder understanding (until you learn topology...).