a) Obliczamy granicę. Wyłączamy przed nawias najwyższą potęgę  $n$  w liczniku i mianowniku.

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

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b) Obliczamy granicę. Wyłączamy przed nawias najwyższą potęgę  $n$  w liczniku i mianowniku.

$\underset{n\to \infty }{\lim }\frac{3n^2+7n}{n^6+2}=\underset{n\to \infty }{\lim }\frac{n^2\left(3+\frac{7}{n}\right)}{n^6\left(1+\frac{2}{n^6}\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{3}{\uparrow }}{\overbrace{3}}+\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{7}{n}}}}{\underset{\underset{+\infty }{\downarrow }}{\underbrace{n^4}}\text{(}\underset{\underset{1}{\downarrow }}{\underbrace{1}}+\underset{\underset{0}{\downarrow }}{\underbrace{\frac{2}{n^6}}}\text{)}}=\frac{3+0}{\left(+\infty \right)\cdot \left(1+0\right)}=\frac{3}{\left(+\infty \right)\cdot 1}=\frac{3}{+\infty }=0$  

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c) Obliczamy granicę. Wyłączamy przed nawias najwyższą potęgę  $n$  w liczniku i mianowniku.

$\underset{n\to \infty }{\lim }\frac{6+10n^5}{4n-3n^2-2n^5}=\underset{n\to \infty }{\lim }\frac{n^5\left(\frac{6}{n^5}+10\right)}{n^5\left(\frac{4}{n^4}-\frac{3}{n^3}-2\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{6}{n^5}}}+\stackrel{\stackrel{10}{\uparrow }}{\overbrace{10}}}{\underset{\underset{0}{\downarrow }}{\underbrace{\frac{4}{n^4}}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{3}{n^3}}}-\underset{\underset{2}{\downarrow }}{\underbrace{2}}}=\frac{0+10}{0-0-2}=\frac{10}{-2}=-5$  

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d) Obliczamy granicę. Wyłączamy przed nawias najwyższą potęgę  $n$  w liczniku i mianowniku.

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

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e) Obliczamy granicę. Wyłączamy przed nawias najwyższą potęgę  $n$  w liczniku i mianowniku.

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

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f) Obliczamy granicę. Wyłączamy przed nawias najwyższą potęgę  $n$  w liczniku i mianowniku.

$\underset{n\to \infty }{\lim }\frac{4-3n^6}{5n^3+n^2-2n+1}=\underset{n\to \infty }{\lim }\frac{n^6\left(\frac{4}{n^6}-3\right)}{n^3\left(5+\frac{1}{n}-\frac{2}{n^2}+\frac{1}{n^3}\right)}=\underset{n\to \infty }{\lim }\frac{\stackrel{\stackrel{+\infty }{\uparrow }}{\overbrace{n^3}}\text{(}\stackrel{\stackrel{0}{\uparrow }}{\overbrace{\frac{4}{n}}}-\stackrel{\stackrel{3}{\uparrow }}{\overbrace{3}}\text{)}}{\underset{\underset{5}{\downarrow }}{\underbrace{5}}+\underset{\underset{0}{\downarrow }}{\underbrace{\frac{1}{n}}}-\underset{\underset{0}{\downarrow }}{\underbrace{\frac{2}{n^2}}}+\underset{\underset{0}{\downarrow }}{\underbrace{\frac{1}{n^3}}}}=\frac{\left(+\infty \right)\cdot \left(0-3\right)}{5+0-0+0}=\frac{\left(+\infty \right)\cdot \left(-3\right)}{5}=\frac{-\infty }{5}=-\infty$