The Limit Laws fail when the denominator tends to 0. If the numerator also tends to 0, the expression has the form $\frac{0}{0}$. That form is not an answer: it is a sign that the top and bottom share a factor. Your factoring skills from Simplifying Rational Expressions are exactly what is needed.
Idea clave
Por qué funciona
Why can't $\frac{0}{0}$ be given a value? Because different $\frac{0}{0}$ forms have different limits. At $x = 0$: $\frac{x}{x} \to 1$, $\frac{x^2}{x} \to 0$, and $\frac{3x}{x} \to 3$, while $\frac{x}{x^2}$ blows up. The form alone does not decide the answer; the shared factor does.
Ejemplo resuelto: Cancel a linear factor
- Substituting gives $\frac{9 - 3 - 6}{0} = \frac{0}{0}$, so $(x - 3)$ must be a factor of the top.
- Factor: $x^2 - x - 6 = (x - 3)(x + 2)$.
- Cancel: for $x \ne 3$ the expression equals $x + 2$.
- Substitute: $3 + 2 = 5$.
Respuesta: $5$
Ejemplo resuelto: Factors on both levels
- Substituting gives $\frac{0}{0}$.
- Factor: $\frac{(x - 2)(x + 2)}{(x - 2)(x - 3)}$.
- Cancel $(x - 2)$: $\frac{x + 2}{x - 3}$.
- Substitute: $\frac{4}{-1} = -4$.
Respuesta: $-4$
Ejemplo resuelto: A difference of cubes
- $\frac{0}{0}$ again. Use $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$.
- Cancel: the expression is $x^2 + 2x + 4$ for $x \ne 2$.
- Substitute: $4 + 4 + 4 = 12$.
Respuesta: $12$ (this will turn out to be the slope of $y = x^3$ at $x = 2$)
Factoring patterns that come up
- Trinomials: if $x = a$ makes the top 0, then $(x - a)$ is a factor; the other factor follows from the constant term.
- Difference of squares: $x^2 - a^2 = (x - a)(x + a)$.
- Cubes: $x^3 - a^3 = (x - a)(x^2 + ax + a^2)$ and $x^3 + a^3 = (x + a)(x^2 - ax + a^2)$.
- Complex fractions: combine the small fractions into one, then look for the factor.
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