Find a zero-denominator limit using factorization Calculus

\lim\limits_{x \to 3}{\dfrac{4x-12}{2x^2-10x+12}}=\infty

\lim\limits_{x \to 3}{\dfrac{4x-12}{2x^2-10x+12}}=2

\lim\limits_{x \to 3}{\dfrac{4x-12}{2x^2-10x+12}}=\dfrac{2}{3}

\lim\limits_{x \to 3}{\dfrac{4x-12}{2x^2-10x+12}}=-1

\lim\limits_{x \rightarrow 3} \dfrac{2x^2 - 4x - 6}{3x^2 - 3x - 18} = \dfrac{8}{15}

\lim\limits_{x \rightarrow 3} \dfrac{2x^2 - 4x - 6}{3x^2 - 3x - 18} = \dfrac{8}{5}

\lim\limits_{x \rightarrow 3} \dfrac{2x^2 - 4x - 6}{3x^2 - 3x - 18} = \dfrac{4}{5}

\lim\limits_{x \rightarrow 3} \dfrac{2x^2 - 4x - 6}{3x^2 - 3x - 18} = \dfrac{2}{3}

\lim\limits_{x \rightarrow \frac{-4}{5}} \dfrac{5x+4}{x^2 + x + \dfrac{4}{25}} = \dfrac{-25}{3}

\lim\limits_{x \rightarrow \frac{-4}{5}} \dfrac{5x+4}{x^2 + x + \dfrac{4}{25}} = -4

\lim\limits_{x \rightarrow \frac{-4}{5}} \dfrac{5x+4}{x^2 + x + \dfrac{4}{25}} = -16

\lim\limits_{x \rightarrow \frac{-4}{5}} \dfrac{5x+4}{x^2 + x + \dfrac{4}{25}} = \dfrac{-3}{5}

\lim\limits_{x \rightarrow \frac{2}{5}} \dfrac{5x - 2}{5x^2 - 27x + 10} = \dfrac{-5}{23}

\lim\limits_{x \rightarrow \frac{2}{5}} \dfrac{5x - 2}{5x^2 - 27x + 10} = \dfrac{-23}{5}

\lim\limits_{x \rightarrow \frac{2}{5}} \dfrac{5x - 2}{5x^2 - 27x + 10} = \dfrac{-3}{5}

\lim\limits_{x \rightarrow \frac{2}{5}} \dfrac{5x - 2}{5x^2 - 27x + 10} = 5

\lim\limits_{x \rightarrow \frac{1}{2}} \dfrac{2x^2 + 3x - 2}{2x - 1} = \dfrac{5}{2}

\lim\limits_{x \rightarrow \frac{1}{2}} \dfrac{2x^2 + 3x - 2}{2x - 1} = \dfrac{5}{4}

\lim\limits_{x \rightarrow \frac{1}{2}} \dfrac{2x^2 + 3x - 2}{2x - 1} = 0

\lim\limits_{x \to 3}{\dfrac{4x-12}{2x^2-10x+12}}=-1

\lim\limits_{x \rightarrow 4} \dfrac{2x^2 - 32}{x^2 - x - 12} = \dfrac{16}{7}

\lim\limits_{x \rightarrow 4} \dfrac{2x^2 - 32}{x^2 - x - 12} = 8

\lim\limits_{x \rightarrow 4} \dfrac{2x^2 - 32}{x^2 - x - 12} = 2

\lim\limits_{x \rightarrow 4} \dfrac{2x^2 - 32}{x^2 - x - 12} = \infty

\lim\limits_{x \rightarrow 2} \dfrac{x^2 - 4}{3x^3 + 6x^2 - 12x - 24} = \dfrac{1}{12}

\lim\limits_{x \rightarrow 2} \dfrac{x^2 - 4}{3x^3 + 6x^2 - 12x - 24} = \dfrac{1}{48}

\lim\limits_{x \rightarrow 2} \dfrac{x^2 - 4}{3x^3 + 6x^2 - 12x - 24} = \dfrac{-4}{3}

\lim\limits_{x \rightarrow 2} \dfrac{x^2 - 4}{3x^3 + 6x^2 - 12x - 24} = \infty