Infinite Limits
Limits that Equal Infinity
Consider the function $f(x) = \frac{1}{(x-3)^2}$.
We currently cannot describe what happens at $x = 3$. This limit does not approach a finite value. Thus, we have to extend the allowable values of $L$ to the extended real numbers \(\{ - \infty \} \cup \mathbb{R} \cup \{ \infty \}\). Now, we have to define what it means for a limit to equal positive or negative infinity. Suppose $a \in \mathbb{R}$.
\[\lim_{x \rightarrow a} f(x) = \infty \qquad \iff \qquad \forall M > 0 \quad \exists \delta > 0 \quad \text{s.t.} \quad 0 < \lvert x - a \rvert < \delta \implies f(x) > M\] \[\lim_{x \rightarrow a} f(x) = -\infty \qquad \iff \qquad \forall M > 0 \quad \exists \delta > 0 \quad \text{s.t.} \quad 0 < \lvert x - a \rvert < \delta \implies f(x) < -M\]Intuitively, this is saying no matter what upperbound I give you, given a point sufficiently close to $a$, the function will surpass this threshold.
Of course, you can break this up into its left- and right-handed limits just as we did before. Often you have to in order to properly analyze functions. For example, $f(x) = \frac{1}{x-3}$.
\[\lim_{x \rightarrow 3^-} \frac{1}{x-3} = -\infty \qquad\qquad \lim_{x \rightarrow 3^+} \frac{1}{x-3} = \infty\]Therefore, $\lim_{x \rightarrow 3} \frac{1}{x-3}$ is undefined (just as before).
Limits that Approach Infinity
In the above, we assumed that \(L \in \{ -\infty, \infty \}\) and $a \in \mathbb{R}$. If the limit exists under these conditions it is called an infinite discontinuity or a vertical asymptote. Here, we are going to assume the opposite, i.e. \(a \in \{ -\infty, \infty \}\) and $L \in \mathbb{R}$. This will create a horizontal asymptote.
\[\lim_{x \rightarrow \infty} f(x) = L \qquad \iff \qquad \forall \epsilon > 0 \quad \exists N > 0 \quad \text{s.t.} \quad x > N \implies 0 < \lvert f(x) - L \rvert < \epsilon\] \[\lim_{x \rightarrow -\infty} f(x) = L \qquad \iff \qquad \forall \epsilon > 0 \quad \exists N > 0 \quad \text{s.t.} \quad x < -N \implies 0 < \lvert f(x) - L \rvert < \epsilon\]This is the same as before but flipped 90 degrees. No matter how close to $L$ we get, I can always find a large enough value of $x$ such that $f(x)$ falls within that range.
All the Infinities
Finally, we can assume that both \(a, L \in \{ -\infty, \infty \}\). There are four combinations.
\[\lim_{x \rightarrow \infty} f(x) = \infty \qquad \iff \qquad \forall M > 0 \quad \exists N > 0 \quad \text{s.t.} \quad x > N \implies f(x) > M\] \[\lim_{x \rightarrow -\infty} f(x) = \infty \qquad \iff \qquad \forall M > 0 \quad \exists N > 0 \quad \text{s.t.} \quad x < -N \implies f(x) > M\] \[\lim_{x \rightarrow \infty} f(x) = -\infty \qquad \iff \qquad \forall M > 0 \quad \exists N > 0 \quad \text{s.t.} \quad x > N \implies f(x) < -M\] \[\lim_{x \rightarrow -\infty} f(x) = -\infty \qquad \iff \qquad \forall M > 0 \quad \exists N > 0 \quad \text{s.t.} \quad x < -N \implies f(x) < -M\]Limit Laws
I am not going to prove it here (because it would take an entire new series), but all of the previous limit laws also hold for these limit definitions.