The section title might make you think we’re reversing course on our original definition, but only somewhat. When a function $f(x)$ is unbounded as $x$ approaches $c$, the limit still does not exist. But we can say also state that the limit is infinity for the sake of stating what’s going on.
\[\lim_{x\to 0^-} \frac{1}{x} = -\infty \qquad \lim_{x\to 0^+} \frac{1}{x} = \infty\]Determining when the limit is infinity it much the same as other limits. Look at it graphically or use values close to $c$. With either method, you are looking for large increases as you get closer to $c$.
When $f(x)$ approaches infinity (or negative infinity), we get imaginary lines called vertical asymptotes. You can spot these without graphing by looking for when the denominator is 0, without the numerator also being 0.
The function
\[f(x) = \frac{1}{2(x+1)}\]has one at $x=-1$, since the denominator will evaluate to 0, but not the numerator.
In the function
\[f(x) = \frac{x-1}{(x-1)(x+1)}\]there is a vertical asymptote again at $x=-1$, but not at $x=1$. At $f(1)$, both the numerator and denominator evaluate to 0, and we have a case of removable discontinuity.
The book has some properties of infinite limits that should be noted. I suggest looking closely at the one for quotients. See if you determine why the result is 0.