Repairing a hole AB & BC

To repair a discontinuity you assign a value at the point that makes the function continuous there. Condition 3 of the continuity test says that value must equal $\lim_{x\to a} f(x)$, so the repair is possible exactly when that limit exists.

That is the definition of a removable discontinuity, which is where the name comes from. For the other types there is nothing to assign:

1. Jump. The one-sided limits differ, so no single value can match both.
2. Infinite. The outputs are unbounded nearby, so no finite value helps.

Naming the type first is therefore not bookkeeping: it is what tells you whether a repair is even available.

For $f(x) = \frac{x^2-9}{x-3}$, the limit at 3 is 6, so define

$$g(x) = \begin{cases}\dfrac{x^2-9}{x-3}, & x \ne 3\\[4pt] 6, & x = 3\end{cases}$$

and $g$ is continuous everywhere. Choosing any other number leaves a discontinuity of exactly the same kind, just with the dot in a different place.

A common wrong instinct is to substitute into the original expression to find the repair value, which returns $\frac{0}{0}$ and no value at all. The repair value comes from the limit, which is what the simplification computes.

Writing $\frac{x^2-9}{x-3} = x+3$ is a valid statement for $x \ne 3$. It does not change the original function, which is still undefined at 3.

What the cancellation does is reveal the limit: since the simplified form is continuous, its value at 3 is the limit of the original. That is a computation, not a repair. The repair is a separate act of defining a new function with a new piece.

So a claim like "after cancelling, $x = 3$ is fine" is false about $f$ and true only about the newly defined $g$. On an exam, if a question asks to remove a discontinuity, writing the simplified expression alone is not a complete answer: the piecewise definition, or an explicit statement of the value assigned, is what was asked for.

For $\frac{x+1}{x-2}$, no factoring makes $x = 2$ acceptable, because nothing cancels: the discontinuity is infinite and the function is genuinely unbounded nearby. Manipulating the algebra cannot change that.

The test is quick. Factor completely and see whether the offending factor cancels. If it does, the discontinuity is removable and the limit is the repair value. If it survives in the denominator, the discontinuity is infinite and no repair exists. That single check separates the two cases and is the same check used in Topic 1.10 to classify them.