Помогите с пределами))))

По правилу Лопиталя
+oo\ \ \frac{5n^3-n^2-4}{3n^3+11n^2+1}=\frac{15n^2-2n}{9n^2+22n}=\frac{30n-2}{18n+22}=\frac{5}{3}" alt="n->+oo\ \ \frac{5n^3-n^2-4}{3n^3+11n^2+1}=\frac{15n^2-2n}{9n^2+22n}=\frac{30n-2}{18n+22}=\frac{5}{3}" align="absmiddle" class="latex-formula"> то есть производные

+oo\ \ \frac{3n^2+4}{n^3+n^2+1}=\frac{6n}{3n^2+2n}=\frac{6}{6n+2}=\frac{6}{+oo}=0\\ " alt="n->+oo\ \ \frac{3n^2+4}{n^3+n^2+1}=\frac{6n}{3n^2+2n}=\frac{6}{6n+2}=\frac{6}{+oo}=0\\ " align="absmiddle" class="latex-formula">

$\sqrt[3]{n+1}-\sqrt[3]{n}=((n+1)^{\frac{1}{9}} - n^{\frac{1}{9}})(n^{\frac{2}{9}}+(n(n+1))^{\frac{1}{9}}+(n+1)^{\frac{2}{9}})=0$

+oo\ \ 4+5n+4n^2-3n^3 \\ n^3(\frac{4}{n^3}+\frac{5}{n^2}+\frac{4}{n}-3)=+oo(0+0+0-3)=-oo" alt="n-->+oo\ \ 4+5n+4n^2-3n^3 \\ n^3(\frac{4}{n^3}+\frac{5}{n^2}+\frac{4}{n}-3)=+oo(0+0+0-3)=-oo" align="absmiddle" class="latex-formula">