Find the limit of a function of the form f/g Calculus

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

\lim\limits_{x \to -2}\dfrac{3+2x}{4^x}=-16

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

\lim\limits_{x \to -2}\dfrac{3+2x}{4^x}=\dfrac{1}{16}

The limit does not exist.

\lim\limits_{x \rightarrow 1} \dfrac{6x - 1}{x^3 - 1} = 0

\lim\limits_{x \rightarrow 1} \dfrac{6x - 1}{x^3 - 1} = 5

\lim\limits_{x \rightarrow 1} \dfrac{6x - 1}{x^3 - 1} = \dfrac{5}{2}

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

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

\lim\limits_{x \rightarrow -1} \dfrac{x^2 + 2x}{x^5 - 1} = 0

The limit does not exist.

\lim\limits_{x \rightarrow 3} \dfrac{3x^2 + x + 1}{6x^2 + 2x} = \dfrac{31}{60}

\lim\limits_{x \rightarrow 3} \dfrac{3x^2 + x + 1}{6x^2 + 2x} = \dfrac{1}{2}

\lim\limits_{x \rightarrow 3} \dfrac{3x^2 + x + 1}{6x^2 + 2x} = \dfrac{3}{2}

\lim\limits_{x \rightarrow 3} \dfrac{3x^2 + x + 1}{6x^2 + 2x} = 2

\lim\limits_{x \rightarrow -2} \dfrac{3^x}{x^3} = \dfrac{-1}{72}

\lim\limits_{x \rightarrow -2} \dfrac{3^x}{x^3} = \dfrac{9}{8}

\lim\limits_{x \rightarrow -2} \dfrac{3^x}{x^3} = 72

\lim\limits_{x \rightarrow -2} \dfrac{3^x}{x^3} = \dfrac{-8}{9}

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

\lim\limits_{x \rightarrow 4} \dfrac{2x - 1}{x^2 - 3^{x-2}} = \infty

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

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

\lim\limits_{x \rightarrow 3} \dfrac{5^x}{-2x^2 + 4x + 1} = -25

\lim\limits_{x \rightarrow 3} \dfrac{5^x}{-2x^2 + 4x + 1} = \dfrac{-125}{7}

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

\lim\limits_{x \rightarrow 3} \dfrac{5^x}{-2x^2 + 4x + 1} = \dfrac{-125}{18}