$\begin{array}{l}c)\quad \lim\dfrac{3n^3 + 2n^2 + n}{n^3 + 4}\\ = \lim\dfrac{3 + \dfrac2n + \dfrac{1}{n^2}}{1 + \dfrac{4}{n^3}}\\ = \dfrac{3 +0+0}{1 + 0}\\ = 3\\ d)\quad \lim\dfrac{n^4}{(n+1)(2+n)(n^2+1)}\\ = \lim\dfrac{1}{\left(1 + \dfrac1n\right)\left(\dfrac2n + 1\right)\left(1 + \dfrac{1}{n^2}\right)}\\ = \dfrac{1}{(1+0)(0+1)(1+0)}\\ = \dfrac13\\ e)\quad \lim\dfrac{n^2+1}{2n^4 + n + 1}\\ = \lim\dfrac{\dfrac{1}{n^2}+ \dfrac{1}{n^4}}{2 + \dfrac{1}{n^3} + \dfrac{1}{n^4}}\\ =\dfrac{0+0}{2+0+0}\\ = 0\\ f)\quad \lim\dfrac{2n^4 + n^2 -3}{3n^3 -2n^2 +1}\\ = \lim\dfrac{n\left(2 + \dfrac{1}{n^2} - \dfrac{3}{n^4}\right)}{3 -\dfrac{2}{n} + \dfrac{1}{n^3}}\\ = \dfrac{+\infty\cdot 2}{3}\\ = +\infty \end{array}$
