Rewriting to resolve a limit AB & BC

A $\frac{0}{0}$ at $x = a$ means both numerator and denominator vanish there, so both contain a factor of $(x - a)$. Cancelling it produces a new expression that agrees with the original at every $x \ne a$ and is defined at $a$ as well.

That is exactly enough, because the limit never looks at $x = a$. Two functions agreeing on a punctured neighborhood have the same limit there, so:

$$\lim_{x\to 3}\frac{x^2-9}{x-3} = \lim_{x\to 3}\frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3}(x+3) = 6.$$

The cancellation is not "allowed because the factor is nonzero"; it is allowed because the two expressions have the same limit. The original is still undefined at 3, and the answer 6 is still correct.

Cancellation removes a factor shared by the whole numerator and the whole denominator. It never removes a term from inside a sum. This is the rule that keeps everything else honest:

1. $\dfrac{x^2+3x}{x} = \dfrac{x(x+3)}{x} = x+3$. Valid: $x$ was factored out first.
2. $\dfrac{x^2+3x}{x} = x^2+3$. Invalid. The $x$ was struck from one term only.
3. $\dfrac{x^2+1}{x^2+5}$ cannot be simplified at all. Nothing is a shared factor.

The test: can you write both numerator and denominator as products, with the cancelled piece appearing as one whole factor of each? If not, the cancellation is invalid, and the resulting limit will simply be wrong rather than merely unsimplified.

Factoring handles polynomials. Difference of squares, common factors, and trinomials cover nearly every case: $\frac{x^2-5x+6}{x-2} = \frac{(x-2)(x-3)}{x-2} \to 1$ as $x \to 2$.

Conjugates handle radicals. Multiply top and bottom by the conjugate to convert a difference of roots into a difference of squares:

$$\frac{\sqrt{x+4}-2}{x} \cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \frac{x}{x\left(\sqrt{x+4}+2\right)} = \frac{1}{\sqrt{x+4}+2} \to \frac{1}{4}.$$

Common denominators handle complex fractions. Combine the small fractions first, then simplify the resulting single fraction. For $\frac{\frac{1}{x}-\frac{1}{3}}{x-3}$, the numerator becomes $\frac{3-x}{3x}$, and the $3-x$ cancels against $x-3$ up to a sign, leaving $-\frac{1}{9}$.

After rewriting, substitute again and classify again. Sometimes one pass is not enough and a second factor cancels. Sometimes the rewritten form gives $\frac{k}{0}$, which means the original had an asymptote rather than a hole all along, and the correct answer is that the limit does not exist.

Watch the sign when the cancelled factors are reversed. $x - 3$ and $3 - x$ differ by a factor of $-1$, and losing it flips the answer's sign, which is the most common arithmetic slip in this topic. Writing $3 - x = -(x-3)$ explicitly costs one line and prevents it.