a) Przykładowe rozwiązania:

$a_n=2n^3+4$  

$b_n=2n^3-n-1$  

Wówczas:

$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\underset{n\to \infty }{\lim }\left(\frac{2n^3+4}{2n^3-n-1}\right)=\underset{n\to \infty }{\lim }\frac{n^3\left(2+\frac{4}{n^3}\right)}{n^3\left(2-\frac{1}{n^2}-\frac{1}{n^3}\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{2}{\uparrow }}{\overbrace{2}}+\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{4}{n^3}}}}{\underset{\underset{2}{\downarrow }}{\underbrace{2}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{1}{n^2}}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{1}{n^3}}}}=\frac{2+0}{2-0-0}=\frac{2}{2}=1$  

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b) Przykładowe rozwiązania:

$a_n=n^4+3n+5$  

$b_n=4n^3-2n^2-1$  

Wówczas:

$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\underset{n\to \infty }{\lim }\left(\frac{n^4+3n+5}{4n^3-2n^2-1}\right)=\underset{n\to \infty }{\lim }\frac{n^4\left(1+\frac{3}{n^3}+\frac{5}{n^4}\right)}{n^3\left(4-\frac{2}{n}-\frac{1}{n^3}\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{+\infty }{\uparrow }}{\overbrace{n}}\cdot \text{(}\stackrel{\stackrel{1}{\uparrow }}{\overbrace{1}}+\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{3}{n^3}}}+\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{5}{n^4}}}\text{)}}{\underset{\underset{4}{\downarrow }}{\underbrace{4}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{2}{n}}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{1}{n^3}}}}=\frac{\left(+\infty \right)\cdot \left(1+0+0\right)}{4-0-0}=\frac{\left(+\infty \right)\cdot 1}{4}=\frac{+\infty }{4}=+\infty$  

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c) Przykładowe rozwiązania:

$a_n=-n^3+2$  

$b_n=2n^2-1$  

Wówczas:

$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\underset{n\to \infty }{\lim }\left(\frac{-n^3+2}{2n^2-1}\right)=\underset{n\to \infty }{\lim }\frac{n^3\left(-1+\frac{2}{n^3}\right)}{n^2\left(2-\frac{1}{n^2}\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{+\infty }{\uparrow }}{\overbrace{n}}\cdot \text{(}\stackrel{\stackrel{-1}{\uparrow }}{\overbrace{-1}}+\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{2}{n^3}}}\text{)}}{\underset{\underset{2}{\downarrow }}{\underbrace{2}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{1}{n^2}}}}=\frac{\left(+\infty \right)\cdot \left(-1+0\right)}{2-0}=\frac{\left(+\infty \right)\cdot \left(-1\right)}{2}=\frac{-\infty }{2}=-\infty$  

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d) Przykładowe rozwiązania:

$a_n=5n$  

$b_n=-n+2$  

Wówczas:

$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\underset{n\to \infty }{\lim }\left(\frac{5n}{-n+2}\right)=\underset{n\to \infty }{\lim }\frac{n\cdot 5}{n\left(-1+\frac{2}{n}\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{5}{\uparrow }}{\overbrace{5}}}{\underset{\underset{-1}{\downarrow }}{\underbrace{-1}}+\underset{\underset{0}{\downarrow }}{\underbrace{\frac{2}{n}}}}=\frac{5}{-1+0}=\frac{5}{-1}=-5$  

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e) Przykładowe rozwiązania:

$a_n=2n-3$  

$b_n=5n^2+6n+2$  

Wówczas:

$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\underset{n\to \infty }{\lim }\left(\frac{2n-3}{5n^2+6n+2}\right)=\underset{n\to \infty }{\lim }\frac{n\left(2-\frac{3}{n}\right)}{n^2\left(5+\frac{6}{n}+\frac{2}{n^2}\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{2}{\uparrow }}{\overbrace{2}}-\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{3}{n}}}}{\underset{\underset{+\infty }{\downarrow }}{\underbrace{n}}\cdot \text{(}\underset{\underset{5}{\downarrow }}{\underbrace{5}}+\underset{\underset{0}{\downarrow }}{\underbrace{\frac{6}{n}}}+\underset{\underset{0}{\downarrow }}{\underbrace{\frac{2}{n^2}}}\text{)}}=\frac{2-0}{\left(+\infty \right)\cdot \left(5+0+0\right)}=\frac{2}{\left(+\infty \right)\cdot 5}=\frac{2}{+\infty }=0$  

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f) Przykładowe rozwiązania:

$a_n=-2n^3+4$  

$b_n=4n^3+n-1$  

Wówczas:

$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\underset{n\to \infty }{\lim }\left(\frac{-2n^3+4}{4n^3+n-1}\right)=\underset{n\to \infty }{\lim }\frac{n^3\left(-2+\frac{4}{n^3}\right)}{n^3\left(4-\frac{1}{n^2}-\frac{1}{n^3}\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{-2}{\uparrow }}{\overbrace{-2}}+\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{4}{n^3}}}}{\underset{\underset{4}{\downarrow }}{\underbrace{4}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{1}{n^2}}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{1}{n^3}}}}=\frac{-2+0}{4-0-0}=\frac{-2}{4}=-\frac{1}{2}$