a)

Obliczmy granicę:  

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

$=\frac{-4+0+0}{0+0+2}=-2$   

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b)

Obliczmy granicę:

$\underset{n\to \infty }{\lim }\frac{10n^2-2}{2-n+n^3}=\underset{n\to \infty }{\lim }\frac{n^3\left(\frac{10}{n}-\frac{2}{n^2}\right)}{n^3\left(\frac{2}{n^3}-\frac{1}{n^2}+1\right)}=\underset{n\to \infty }{\lim }\frac{\frac{10}{n}-\frac{2}{n^2}}{\frac{2}{n^3}-\frac{1}{n^2}+1}=\frac{0-0}{0+0+1}=0$  

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c)

Obliczmy granicę:

$\underset{n\to \infty }{\lim }\frac{6n^4+n^2-5}{2-n^3}=\underset{n\to \infty }{\lim }\frac{n^3\left(6n+\frac{1}{n}-\frac{5}{n^3}\right)}{n^3\left(\frac{2}{n^3}-1\right)}=\underset{n\to \infty }{\lim }\frac{6n+\frac{1}{n}-\frac{5}{n^3}}{\frac{2}{n^3}-1}=\left[\frac{\infty }{-1}\right]=-\infty$  

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d)

Obliczmy granicę:

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

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e)

Obliczmy granicę:

$\underset{n\to \infty }{\lim }\frac{6-2n-9n^2}{6+2n+3n^2}=\underset{n\to \infty }{\lim }\frac{n^2\left(\frac{6}{n^2}-\frac{2}{n}-9\right)}{n^2\left(\frac{6}{n^2}+\frac{2}{n}+3\right)}=\underset{n\to \infty }{\lim }\frac{\frac{6}{n^2}-\frac{2}{n}-9}{\frac{6}{n^2}+\frac{2}{n}+3}=\frac{0-0-9}{0+0+3}=-3$  

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f)

Obliczmy granicę:

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

$=\frac{-3+0+0}{4+0+0}=-\frac{3}{4}$