Determining Limits Using Algebraic Manipulation

Getting 0/0 when you substitute means direct substitution failed, not that the limit doesn't exist. A limit only cares about what the function does near a point, not what happens exactly at it. If both the top and bottom hit zero at the same x-value, they share a common factor, and that factor is creating a hole in the graph right where you're trying to evaluate. The algebra's job is to expose and remove that shared factor so you can see through the hole to the value the function approaches.

Factoring does this directly: write the numerator and denominator as products, spot the factor they have in common, and cancel it. This is legal everywhere except at the exact point causing the hole, which is fine, because a limit never needs the function to be defined there anyway. Once the problem factor is gone, substitution works on what's left. For (x²-9)/(x-3) as x approaches 3, factor the top as (x-3)(x+3), cancel (x-3), and you're left with x+3, which gives 6. For (x²-5x+6)/(x²-4) as x approaches 2, factor top and bottom as (x-2)(x-3) and (x-2)(x+2), cancel (x-2), and substitution gives (2-3)/(2+2) = -1/4.

This technique is the main tool for evaluating limits that resist direct substitution, and it sets up the next idea: some functions still won't simplify this way, which is when rationalizing conjugates or other tricks take over. It also previews continuity, since a function with a removable hole is discontinuous there even though its limit exists.