$\text{a)}\underset{n\to \infty }{\lim }\frac{n}{n^2+3n}=\underset{n\to \infty }{\lim }\frac{n^2\left(\frac{1}{n}\right)}{n^2\left(1+\frac{3}{n}\right)}=\underset{n\to \infty }{\lim }\frac{\frac{1}{n}}{1+\frac{3}{n}}=\frac{0}{1}=0$       

$\text{b)}\underset{n\to \infty }{\lim }\frac{-5n^2+3n-1}{2n^2+3}=\underset{n\to \infty }{\lim }\frac{n^2\left(-5+\frac{3}{n}-\frac{1}{n^2}\right)}{n^2\left(2+\frac{3}{n^2}\right)}=\underset{n\to \infty }{\lim }\frac{-5+\frac{3}{n}-\frac{1}{n^2}}{2+\frac{3}{n^2}}=-\frac{5}{2}=-2\frac{1}{2}$  

$\text{c)}\underset{n\to \infty }{\lim }\frac{4n^2-n+1}{3n-n^2}=\underset{n\to \infty }{\lim }\frac{n^2\left(4-\frac{1}{n}+\frac{1}{n^2}\right)}{n^2\left(\frac{3}{n}-1\right)}=\underset{n\to \infty }{\lim }\frac{4-\frac{1}{n}+\frac{1}{n^2}}{\frac{3}{n}-1}=\frac{4}{-1}=-4$  

$\text{d)}\underset{n\to \infty }{\lim }\frac{3n^2-7n-2}{15n^2+6n-5}=\underset{n\to \infty }{\lim }\frac{n^2\left(3-\frac{7}{n}-\frac{2}{n^2}\right)}{n^2\left(15+\frac{6}{n}-\frac{5}{n^2}\right)}=\underset{n\to \infty }{\lim }\frac{3-\frac{7}{n}-\frac{2}{n^2}}{15+\frac{6}{n}-\frac{5}{n^2}}=\frac{3}{15}=\frac{1}{5}$  

$\text{e)}\underset{n\to \infty }{\lim }\frac{-7n^3+8n^2-9n+1}{5-4n+3n^2-n^3}=\underset{n\to \infty }{\lim }\frac{n^3\left(-7+\frac{8}{n}-\frac{9}{n^2}+\frac{1}{n^3}\right)}{n^3\left(\frac{5}{n^2}-\frac{4}{n^2}+\frac{3}{n}-1\right)}=$  

$=\underset{n\to \infty }{\lim }\frac{-7+\frac{8}{n}-\frac{9}{n^2}+\frac{1}{n^3}}{\frac{5}{n^2}-\frac{4}{n^2}+\frac{3}{n}-1}=\frac{-7}{-1}=7$  

$\text{f)}\underset{n\to \infty }{\lim }\frac{\left(1+n\right)\left(5n+3\right)}{n\left(3-n\right)}=\underset{n\to \infty }{\lim }\frac{5n+3+5n^2+3n}{3n-n^2}=\underset{n\to \infty }{\lim }\frac{5n^2+8n+3}{3n-n^2}=$  

$=\underset{n\to \infty }{\lim }\frac{n^2\left(5+\frac{8}{n}+\frac{3}{n^2}\right)}{n^2\left(\frac{3}{n}-1\right)}=\underset{n\to \infty }{\lim }\frac{5+\frac{8}{n}+\frac{3}{n^2}}{\frac{3}{n}-1}=\frac{5}{-1}=-5$  

$\text{g)}\underset{n\to \infty }{\lim }\frac{4n^2-n+3}{4-3n+8n^2}=\underset{n\to \infty }{\lim }\frac{n^2\left(4-\frac{1}{n}+\frac{3}{n^2}\right)}{n^2\left(\frac{4}{n^2}-\frac{3}{n}+8\right)}=\underset{n\to \infty }{\lim }\frac{4-\frac{1}{n}+\frac{3}{n^2}}{\frac{4}{n^2}-\frac{3}{n}+8}=\frac{4}{8}=\frac{1}{2}$  

$\text{h)}\underset{n\to \infty }{\lim }\frac{3n^4+8n-5}{9n^4-7n^2+2n}=\underset{n\to \infty }{\lim }\frac{n^4\left(3+\frac{8}{n^3}-\frac{5}{n^4}\right)}{n^4\left(9-\frac{7}{n^2}+\frac{2}{n^3}\right)}=\underset{n\to \infty }{\lim }\frac{3+\frac{8}{n^3}-\frac{5}{n^4}}{9-\frac{7}{n^2}+\frac{2}{n^3}}=\frac{3}{9}=\frac{1}{3}$  

$\text{i)}\underset{n\to \infty }{\lim }\frac{8-5n^3}{3n+11n^3}=\underset{n\to \infty }{\lim }\frac{n^3\left(\frac{8}{n^3}-5\right)}{n^3\left(\frac{3}{n^2}+11\right)}=\underset{n\to \infty }{\lim }\frac{\frac{8}{n^3}-5}{\frac{3}{n^2}+11}=\frac{-5}{11}$