$\text{Jez˙eli}\underset{n\to \infty }{\lim }{a}_n=a,\underset{n\to \infty }{\lim }{b}_n=b,a,b\in \mathbb{R},\text{to:}$ $\underset{n\to \infty }{\lim }\left(c\cdot a_n\right)=c\cdot a,c\in \mathbb{R}$ $\underset{n\to \infty }{\lim }\left(a_n+b_n\right)=\underset{n\to \infty }{\lim }{a}_n+\underset{n\to \infty }{\lim }{b}_n=a+b$ $\underset{n\to \infty }{\lim }\left(a_n-b_n\right)=\underset{n\to \infty }{\lim }{a}_n-\underset{n\to \infty }{\lim }{b}_n=a-b$

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a)

$a_n=\frac{-2n^2+4}{n^2},n\in {\mathbb{N}}_{+}$

Obliczamy granicę ciągu:

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{-2n^2+4}{n^2}=$

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

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

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

$=-2+0=-2$

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b)

$a_n=\frac{6n^3-4n^2+n}{2n^3},n\in {\mathbb{N}}_{+}$

Obliczamy granicę ciągu:

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{6n^3-4n^2+n}{2n^3}=$

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

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

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

$=3-0+\frac{1}{2}\cdot 0=3$

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c)

$a_n=\frac{{\left(n+1\right)}^2}{n^2},n\in {\mathbb{N}}_{+}$

Obliczamy granicę ciągu:

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

$=\underset{n\to \infty }{\lim }\frac{n^2+2n+1}{n^2}=$

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

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

$=\underset{n\to \infty }{\lim }1+\underset{n\to \infty }{\lim }\frac{2}{n}+\underset{n\to \infty }{\lim }\frac{1}{n^2}=$

$=1+0+0=1$