a)

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

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b)

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

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c)

$\underset{n\to \infty }{\lim }\frac{\left(1-3n\right)\left(5-2n\right)}{2n^2+1}=\underset{n\to \infty }{\lim }\frac{5-2n-15n+6n^2}{2n^2+1}=\underset{n\to \infty }{\lim }\frac{6n^2-17n+5}{2n^2+1}=$  

$=\underset{n\to \infty }{\lim }\frac{n^2\left(6-\frac{17}{n}+\frac{5}{n^2}\right)}{n^2\left(2+\frac{1}{n^2}\right)}=\underset{n\to \infty }{\lim }\frac{6-\frac{17}{n}+\frac{5}{n^2}}{2+\frac{1}{n^2}}=\frac{6}{2}=3$  

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d)

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

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e)

$\underset{n\to \infty }{\lim }\frac{3n^4+8n^2-10}{\left(2n^2-1\right)\left(2n^2+1\right)}=\underset{n\to \infty }{\lim }\frac{3n^4+8n^2-10}{4n^4-1}=\underset{n\to \infty }{\lim }\frac{n^4\left(3+\frac{8}{n^2}-\frac{10}{n^4}\right)}{n^4\left(4-\frac{1}{n^4}\right)}=$  

$=\underset{n\to \infty }{\lim }\frac{3+\frac{8}{n^2}-\frac{10}{n^4}}{4-\frac{1}{n^4}}=\frac{3}{4}$  

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f)

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

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