Method of Factorization

P(x) and Q(x) are polynomials in x, and $f(x)=P(x)/Q(x)$. To evaluate $\lim_{x\to a}f(x)$: (1) if $\lim_{x\to a}Q(x)=m\ne0$, the limit is simply $\lim_{x\to a}[P(x)/m]$ — an ordinary substitution, no factoring needed; (2) if $\lim_{x\to a}Q(x)=0$ then $(x-a)$ divides $Q(x)$; if $(x-a)$ does NOT also divide $P(x)$, the limit does not exist (the function is unbounded near a); (3) if $\lim_{x\to a}P(x)$ is also 0, then $(x-a)$ is a genuine common factor of both, and $\lim_{x\to a}f(x)=\lim_{x\to a}\dfrac{P(x)/(x-a)}{Q(x)/(x-a)}$ — factor both polynomials completely, cancel the shared $(x-a)$, and substitute into what remains. Factorizing polynomials this way is the standard tool for rational-function limits that land on the indeterminate form 0/0.\n\nWorked pattern 1 (spotting the common root by inspection): for $\lim_{z\to3}\dfrac{z(2z-3)-9}{z^2-4z+3}$, substituting $z=3$ makes both $z(2z-3)-9$ and $z^2-4z+3$ vanish, so $(z-3)$ is a common factor; factoring gives $\dfrac{(z-3)(2z+3)}{(z-3)(z-1)}$, which cancels to $\dfrac{2z+3}{z-1}$, giving $\dfrac{2(3)+3}{3-1}=\dfrac{9}{2}$ at $z=3$.\n\nWorked pattern 2 (repeated/higher-power factors): for $\lim_{x\to4}\dfrac{(x^3-8x^2+16x)^9}{(x^2-x-12)^{18}}$, the numerator's base factors as $x(x-4)^2$ and the denominator's base as $(x-4)(x+3)$, so raised to the 9th and 18th powers respectively the $(x-4)^{18}$ in top and bottom cancel completely, leaving $\dfrac{x^9}{(x+3)^{18}}\to\dfrac{4^9}{7^{18}}$.\n\nWorked pattern 3 (combining fractions first): for $\lim_{x\to1}\left[\dfrac{1}{x-1}+\dfrac{2}{1-x^2}\right]$, rewriting $1-x^2=(1-x)(1+x)=-(x-1)(x+1)$ and combining over a common denominator produces $\dfrac{(x+1)-2}{(x-1)(x+1)}=\dfrac{x-1}{(x-1)(x+1)}=\dfrac{1}{x+1}\to\dfrac12$.\n\nWorked pattern 4 (synthetic division when the factor isn't obvious by eye): for $\lim_{x\to1}\dfrac{x^3+x^2-5x+3}{x^2-1}$, synthetic division of the numerator by $(x-1)$ gives quotient $x^2+2x-3$, so the numerator is $(x-1)(x^2+2x-3)$ while the denominator factors as $(x-1)(x+1)$; cancelling le …