$\mathop {\lim }\limits_{n \to + \infty}\frac{3n^{3}-5n^{2}+1}{2n^{3}+6n^{2}+4n+5}= \mathop {\lim }\limits_{n \to+ \infty}\frac{n^{3}.(3-\frac{5}{n}+\frac{1}{n^{3}})}{n^{3}.(2+\frac{6}{n}+\frac{4}{n^{2}}+\frac{5}{n^{3}})} = \frac{3}{2}$ $\mathop {\lim }\limits_{n \to + \infty}(\sqrt{n^{2}+3n}-\sqrt{n^{2}}) =\mathop {\lim }\limits_{n \to+\infty}\frac{3n}{n.(\sqrt{1+\frac{3}{n}}+1)} = \frac{3}{2}$
$\mathop {\lim }\limits_{n \to+\infty}\frac{\sqrt{2n^{2}+1}+5n}{1-3n^{2}} = \mathop {\lim }\limits_{n \to +\infty}\frac{n^{2}.(\sqrt{\frac{2}{n^{2}}+\frac{1}{n^{4}}}+\frac{5}{n})}{n^{2}.(\frac{1}{n^{2}}-3)} = 0$