$8)\displaystyle\lim_{x \to 1} \dfrac{x^3-1}{x^2-1}\\ =\displaystyle\lim_{x \to 1} \dfrac{(x-1)(x^2+x+1)}{(x-1)(x+1)}\\ =\displaystyle\lim_{x \to 1} \dfrac{x^2+x+1}{x+1}\\ =\dfrac{1^2+1+1}{1+1}\\ =\dfrac{3}{2}\\ S=a+b=5\\ \Rightarrow B\\ 9)\displaystyle\lim_{x \to 1} \dfrac{x^3+3x-4}{x^2+3x-4}\\ =\displaystyle\lim_{x \to 1} \dfrac{x^3-x^2+x^2-x+4x-4}{x^2-x+4x-4}\\ =\displaystyle\lim_{x \to 1} \dfrac{x^2(x-1)+x(x-1)+4(x-1)}{x(x-1)+4(x-1)}\\ =\displaystyle\lim_{x \to 1} \dfrac{(x-1)(x^2+x+4)}{(x-1)(x+4)}\\ =\displaystyle\lim_{x \to 1} \dfrac{x^2+x+4}{x+4}\\ =\dfrac{1^2+1+4}{1+4}\\ =\dfrac{6}{5}\\ a.b=30\\ \Rightarrow D$
