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

This limit doesn't exist.

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

\lim\limits_{x \to -1} 4^x+3x^2=3.25

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

\lim\limits_{x \rightarrow 2} 4x^3 + \dfrac{2}{x} = 33

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

\lim\limits_{x \rightarrow 2} 4x^3 + \dfrac{2}{x} = 34

This limit doesn't exist.

\lim\limits_{x \rightarrow 3} x^{-2} + x^{2} = \dfrac{82}{9}

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

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

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

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

\lim\limits_{x \rightarrow \infty } \dfrac{3x}{2x^2} + \dfrac{2x}{3} = 0

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

This limit doesn't exist.

The limit doesn't exist.

\lim\limits_{x \rightarrow 4} \dfrac{5x}{4 - x} + 16x^2 - 4x = 240

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

\lim\limits_{x \rightarrow 4} \dfrac{5x}{4 - x} + 16x^2 - 4x = 260

\lim\limits_{x \rightarrow \infty } 3^{-x} + \dfrac{2x - 1}{4x^2} = 0

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

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

This limit doesn't exist.

\lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} + 4x^2 = -126

\lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} + 4x^2 = -198

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

\lim\limits_{x \rightarrow -3} \dfrac{2x}{3^x} + 4x^2 = -90