Three ways a graph can break AB & BC

At a point of discontinuity $x = a$, compute $\lim_{x\to a^-}$ and $\lim_{x\to a^+}$ and compare:

1. Removable (a hole). Both sides exist and are equal, but $f(a)$ is missing or different. The two-sided limit exists.
2. Jump. Both sides exist and are finite but unequal. The two-sided limit does not exist.
3. Infinite. At least one side is unbounded. The two-sided limit does not exist.

This classification is decided entirely by the limits, which is why it is reliable. A fourth case exists, oscillating discontinuity such as $\sin\left(\frac{1}{x}\right)$ at 0, where neither one-sided limit exists at all; the AP course mentions it rarely but it is not removable, a jump, or infinite.

For a rational function, factor first. Then each zero of the original denominator falls into one of two cases:

1. The factor cancels against one in the numerator, leaving a hole.
2. The factor survives in the denominator, giving a vertical asymptote.

Compare $\frac{(x-2)(x+1)}{x-2}$ with $\frac{x+1}{x-2}$. Both have a denominator vanishing at 2. The first has a hole at $(2, 3)$; the second has a vertical asymptote. Nothing but factoring distinguishes them.

Jumps come from a different source: piecewise definitions, and absolute values, which are piecewise in disguise. A rational function without a piecewise definition does not jump.

The rule "denominator zero means vertical asymptote" is the most common wrong shortcut in this unit. It is true only when the factor does not cancel.

Consider $\frac{x^2-9}{x-3}$ at $x = 3$. The denominator vanishes, yet the function equals $x + 3$ everywhere else and the graph is a line with a hole at $(3, 6)$. There is no blow-up anywhere.

The reliable procedure is short: factor completely, cancel what cancels, then look at what remains in the denominator. Zeros of the reduced denominator are the asymptotes; zeros that were cancelled away are the holes. Skipping the factoring step means guessing.

The type determines what can be done about it. A removable discontinuity can be repaired by defining $f(a)$ to be the limit, which is Topic 1.13. Jumps and infinite discontinuities cannot be repaired by any single redefinition, because there is no limit to assign.

The type also controls which theorems apply later. The Intermediate Value Theorem needs continuity on a closed interval, so an asymptote inside the interval disqualifies it, while a removable discontinuity disqualifies it just as firmly until it is repaired. Naming the type is what tells you which of those situations you are in.