$a)\underset{n\to \infty }{\lim }\frac{n\left(6-\frac{1}{n}\right)\cdot n\left(4+\frac{2}{n}\right)}{n\left(2+\frac{1}{n}\right)\cdot n\left(3-\frac{1}{n}\right)}=\frac{6\cdot 4}{2\cdot 3}=\frac{24}{6}=4$  

 

$b)\underset{n\to \infty }{\lim }\frac{n\left(\frac{2}{n}-5\right)\cdot n\left(3+\frac{1}{n}\right)}{n\left(\frac{2}{n}-10\right)\cdot n\left(1+\frac{1}{n}\right)}=\frac{-5\cdot 3}{-10\cdot 1}=\frac{-15}{-10}=1,5$  

 

$c)\underset{n\to \infty }{\lim }\frac{{\left(2+5n\right)}^2}{{\left(n+1\right)}^2}=\underset{n\to \infty }{\lim }{\left(\frac{5n+2}{n+1}\right)}^2=\underset{n\to \infty }{\lim }{\left(\frac{n\left(5+\frac{2}{n}\right)}{n\left(1+\frac{1}{n}\right)}\right)}^2=5^2=25$  

 

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