Jeśli 

$\underset{n\to \infty }{\lim }{a}_n=a,\underset{n\to \infty }{\lim }{b}_n=b$  

to zachodzi 

$\underset{n\to \infty }{\lim }\left(c\cdot a_n\right)=c\cdot a,\text{gdzie }c\in \mathbb{R}$

$\underset{n\to \infty }{\lim }\left(a_n+b_n\right)=\underset{n\to \infty }{\lim }{a}_n+\underset{n\to \infty }{\lim }{b}_n=a+b$

$\underset{n\to \infty }{\lim }\left(a_n-b_n\right)=\underset{n\to \infty }{\lim }{a}_n-\underset{n\to \infty }{\lim }{b}_n=a-b$

$\underset{n\to \infty }{\lim }\left(a_n\cdot b_n\right)=\underset{n\to \infty }{\lim }{a}_n\cdot \underset{n\to \infty }{\lim }{b}_n=a-b$

$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\frac{\underset{n\to \infty }{\lim }{a}_n}{\underset{n\to \infty }{\lim }{b}_n}=\frac{a}{b}$

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$\text{a) }\underset{n\to \infty }{\lim }\frac{2}{n^2}=0$  

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$\text{b) }\underset{n\to \infty }{\lim }\left(5-\frac{2}{n^2}\right)=\underset{n\to \infty }{\lim }5-\underset{n\to \infty }{\lim }\frac{2}{n^2}=5-0=5$  

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$\text{c) }\underset{n\to \infty }{\lim }\left(-2+\frac{1}{\sqrt{n}}\right)=\underset{n\to \infty }{\lim }-2+\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n}}=-2+0=-2$  

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$\text{d) }\underset{n\to \infty }{\lim }\left(4+\frac{1}{n^3}\right)=\underset{n\to \infty }{\lim }4+\underset{n\to \infty }{\lim }\frac{1}{n^3}=4+0=4$  

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

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f)    Rozpatrzmy dwa podciągi:

$a_{2n}=\frac{{\left(-1\right)}^{2n}+{\left(-1\right)}^{2n+1}}{4n}=\frac{1-1}{4n}=\frac{0}{4n}=0$   

$a_{2n-1}=\frac{{\left(-1\right)}^{2n-1}+{\left(-1\right)}^{2n-1+1}}{2\left(2n-1\right)}=\frac{-1+1}{2\left(2n-1\right)}=\frac{0}{2\left(2n-1\right)}=0$  

A więc ciąg (a_n) jest stale równy 0. Zatem:

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }0=0$