$ 9)\displaystyle\lim \dfrac{n(2n+1)(3n^2+2)}{(4n-3)^4}\\ =\displaystyle\lim \dfrac{1\left(2+\dfrac{1}{n}\right)\left(3+\dfrac{2}{n^2}\right)}{\left(4-\dfrac{3}{n}\right)^4}\\ =\dfrac{1.2.3}{4^4}\\ =\dfrac{3}{128}\\ 12)\displaystyle\lim \dfrac{(6-8n)(1-n)^2}{n^4+12}\\ =\displaystyle\lim \dfrac{\dfrac{1}{n}\left(\dfrac{6}{n}-8\right)\left(\dfrac{1}{n}-1\right)^2}{1+\dfrac{12}{n^4}}\\ =\dfrac{0}{1}\\ =0$
