$a)$  

$a_n=\frac{\frac{1}{2}n+8}{2n+4}$  

$\underset{n\to \infty }{\lim }\frac{\frac{1}{2}n+8}{2n+4}=\underset{n\to \infty }{\lim }\frac{\frac{1}{2}+\frac{8}{n}}{2+\frac{4}{n}}=\frac{\frac{1}{2}}{2}=\frac{1}{4}$      

$b)$  

$a_n=\frac{3n^2-12}{2n^2+3n}$

$\underset{n\to \infty }{\lim }\frac{3n^2-12}{2n^2+3n}=\underset{n\to \infty }{\lim }\frac{3-\frac{12}{n^2}}{2+\frac{3}{n}}=\frac{3}{2}$    

$c)$  

$a_n=\frac{-6n^2+n-5}{1+n+3n^2}$  

$\underset{n\to \infty }{\lim }\frac{-6n^2+n-5}{1+n+3n^2}=\underset{n\to \infty }{\lim }\frac{-6+\frac{1}{n}-\frac{5}{n^2}}{\frac{1}{n^2}+\frac{1}{n}+3}=-\frac{6}{3}=-2$     

$d)$  

$a_n=\frac{100n^2+1}{n^4+3n}$  

$\underset{n\to \infty }{\lim }\frac{100n^2+1}{n^4+3n}=\underset{n\to \infty }{\lim }\frac{\frac{100}{n^2}+\frac{1}{n^4}}{1+\frac{3}{n^3}}=\frac{0}{1}=0$   

$e)$  

$a_n=\frac{-4n^3+4n^2+2n-1}{-6n^4+4n^2+1}$  

$\underset{n\to \infty }{\lim }\frac{-4n^3+4n^2+2n-1}{-6n^4+4n^2+1}=\underset{n\to \infty }{\lim }\frac{-\frac{4}{n}+\frac{4}{n^2}+\frac{2}{n^3}-\frac{1}{n^4}}{-6+\frac{4}{n^2}+\frac{1}{n^4}}=\frac{0}{-6}=0$    

$f)$  

$a_n=\frac{8n^8+6n^6-4n^4}{n^5+n^7-n^9}$  

$\underset{n\to \infty }{\lim }\frac{8n^8+6n^6-4n^4}{n^5+n^7-n^9}=\underset{n\to \infty }{\lim }\frac{n^9\left(\frac{8}{n}+\frac{6}{n^3}-\frac{4}{n^5}\right)}{n^9\left(\frac{1}{n^4}+\frac{1}{n^2}-1\right)}=\underset{n\to \infty }{\lim }\frac{\frac{8}{n}+\frac{6}{n^3}-\frac{4}{n^5}}{\frac{1}{n^4}+\frac{1}{n^2}-1}=\frac{0}{-1}=0$