$\text{a)}$  

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

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

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$\text{b)}$  

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(10n^3-\frac{n^4}{10}\right)=\underset{n\to \infty }{\lim }\frac{100n^3-n^4}{10}=$   

$=\underset{n\to \infty }{\lim }\frac{n^4\cdot \left(\frac{100}{n}-1\right)}{10}=\underset{n\to \infty }{\lim }\left(n^4\cdot \frac{\frac{100}{n}-1}{10}\right)\stackrel{\left[+\infty \cdot \left(-\frac{1}{10}\right)\right]}{=}-\infty$  

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$\text{c)}$  

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left({\left(\frac{2}{3}\right)}^n+{\left(\frac{3}{4}\right)}^n-{\left(\frac{5}{4}\right)}^n\right)=$  

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

$=\underset{n\to \infty }{\lim }\frac{8^n+9^n-{15}^n}{{12}^n}=\underset{n\to \infty }{\lim }\frac{{15}^n\cdot \left(\frac{1}{8^n}+\frac{1}{9^n}-1\right)}{{12}^n}=$  

$=\underset{n\to \infty }{\lim }\left({\left(\frac{15}{12}\right)}^n\cdot \left(\frac{1}{8^n}+\frac{1}{9^n}-1\right)\right)\stackrel{\left[\infty \cdot \left(-1\right)\right]}{=}-\infty$  

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$\text{d)}$  

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(\sqrt{n^2+2n+3}-n\right)=\underset{n\to \infty }{\lim }\left(\sqrt{n^2+2n+3}-n\right)\cdot \frac{\sqrt{n^2+2n+3}+n}{\sqrt{n^2+2n+3}+n}=$  

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

$=\underset{n\to \infty }{\lim }\frac{n^2+2n+3-n^2}{\left|n\right|\sqrt{1+\frac{2}{n}+\frac{3}{n^2}}+n}=\underset{n\to \infty }{\lim }\frac{2n+3}{n\sqrt{1+\frac{2}{n}+\frac{3}{n^2}}+n}=$  

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