$\begin{array}{l}g)\quad \lim\limits_{x\to -2}\dfrac{2x^2 + x - 6}{x^3 + 8}\\ = \lim\limits_{x\to -2}\dfrac{(2x-3)(x+2)}{(x+2)(x^2 - 2x + 4)}\\ =\lim\limits_{x\to -2}\dfrac{2x-3}{x^2 - 2x+4}\\ = \dfrac{2.(-2) -3}{(-2)^2 - 2.(-2) + 4}\\ =-\dfrac{7}{12}\\ h)\quad \lim\limits_{x\to 3}\dfrac{x^4 - x^2 -72}{x^2 -2x - 3}\\ = \lim\limits_{x\to 3}\dfrac{(x-3)(x+3)(x^2+8)}{(x-3)(x+1)}\\ = \lim\limits_{x\to 3}\dfrac{(x+3)(x^2+8)}{x+1}\\ = \dfrac{(3+3)(3^2+8)}{3+1}\\ =\dfrac{153}{4} \end{array}$
