$\text{a)}\underset{n\to \infty }{\lim }\frac{6n-31}{n+5}=\underset{n\to \infty }{\lim }\frac{6-\frac{31}{n}}{1+\frac{5}{n}}=\frac{6}{1}=6$      

$\text{b)}\underset{n\to \infty }{\lim }\frac{n-n^2}{n^2+1}=\underset{n\to \infty }{\lim }\frac{\frac{1}{n}-1}{1+\frac{1}{n^2}}=\frac{-1}{1}=-1$  

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

$\text{d)}\underset{n\to \infty }{\lim }\frac{3n^3-1}{n^4-5n+75}=\underset{n\to \infty }{\lim }\frac{\frac{3}{n}-\frac{1}{n^4}}{1-\frac{5}{n^3}+\frac{75}{n^4}}=\frac{0}{1}=0$  

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

$\text{f)}\underset{n\to \infty }{\lim }\frac{n^4}{n^2+5-8n^4}=\underset{n\to \infty }{\lim }\frac{1}{\frac{1}{n^2}+\frac{5}{n^4}-8}=\frac{1}{-8}=-\frac{1}{8}$  

$\text{g)}\underset{n\to \infty }{\lim }\frac{2n}{n^3+5}=\underset{n\to \infty }{\lim }\frac{\frac{2}{n^2}}{1+\frac{5}{n^3}}=\frac{0}{1}=0$  

$\text{h)}\underset{n\to \infty }{\lim }\frac{n^2-6n+100}{7n^2+5}=\underset{n\to \infty }{\lim }\frac{1-\frac{6}{n}+\frac{100}{n^2}}{7+\frac{5}{n^2}}=\frac{1}{7}$  

$\text{i)}\underset{n\to \infty }{\lim }\frac{15-8n^4}{2n^4+n^2+1}=\underset{n\to \infty }{\lim }\frac{\frac{15}{n^4}-8}{2+\frac{1}{n^2}+\frac{1}{n^4}}=\frac{-8}{2}=-4$