Przypomnijmy, że jeśli a jest dowolną liczbą rzeczywistą, to nieskończony ciąg (an) o wyrazie ogólnym $a_n=\frac{a}{n}$   jest zbieżny do 0.

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a)

Zauważmy, że 

• $\underset{n\to \infty }{\lim }\left(5+\frac{3}{n^2}\right)=5+\underset{n\to \infty }{\lim }\left(\frac{3}{n^2}\right)=5+\underset{n\to \infty }{\lim }\frac{3}{n}\cdot \underset{n\to \infty }{\lim }\frac{1}{n}=5+0\cdot 0=5$  
  
  
• $\underset{n\to \infty }{\lim }\frac{n+3}{4n-1}=\underset{n\to \infty }{\lim }\frac{n\left(1+\frac{3}{n}\right)}{n\left(4-\frac{1}{n}\right)}=\underset{n\to \infty }{\lim }\frac{1+\frac{3}{n}}{4-\frac{1}{n}}=\frac{1+0}{4-0}=\frac{1}{4}$  

Zatem mamy 

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

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b)

Obliczamy podaną granicę i otrzymujemy 

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

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

$=\frac{3}{2+0}-\frac{4+0}{6}=\frac{3}{2}-\frac{4}{6}=\frac{9}{6}-\frac{4}{6}=\frac{5}{6}$  

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c)

Zauważmy, że 

• $\underset{n\to \infty }{\lim }\frac{n+1}{2n+3}=\underset{n\to \infty }{\lim }\frac{n\left(1+\frac{1}{n}\right)}{n\left(2+\frac{3}{n}\right)}=\underset{n\to \infty }{\lim }\frac{1+\frac{1}{n}}{2+\frac{3}{n}}=\frac{1+0}{2+0}=\frac{1}{2}$  
   
• $\underset{n\to \infty }{\lim }\frac{2n-5}{n}=\underset{n\to \infty }{\lim }\frac{n\left(2-\frac{5}{n}\right)}{n}=\underset{n\to \infty }{\lim }\frac{2-\frac{5}{n}}{1}=\frac{2-0}{1}=2$  

Zatem mamy

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

$={\left(\underset{n\to \infty }{\lim }\frac{n+1}{2n+3}+\underset{n\to \infty }{\lim }\frac{2n-5}{n}\right)}^2={\left(\frac{1}{2}+2\right)}^2={\left(\frac{5}{2}\right)}^2=\frac{25}{4}=6\frac{1}{4}$  

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d)

Zauważmy, że 

• $\underset{n\to \infty }{\lim }\frac{1-4n}{2n}=\underset{n\to \infty }{\lim }\frac{n\left(\frac{1}{n}-4\right)}{2n}=\underset{n\to \infty }{\lim }\frac{\frac{1}{n}-4}{2}=\frac{0-4}{2}=-2$  
• $\underset{n\to \infty }{\lim }\frac{n}{n+5}=\underset{n\to \infty }{\lim }\frac{n}{n\left(1+\frac{5}{n}\right)}=\underset{n\to \infty }{\lim }\frac{1}{1+\frac{5}{n}}=\frac{1}{1+0}=1$  
• $\underset{n\to \infty }{\lim }\frac{2n+4}{6n+7}=\underset{n\to \infty }{\lim }\frac{n\left(2+\frac{4}{n}\right)}{n\left(6+\frac{7}{n}\right)}=\underset{n\to \infty }{\lim }\frac{2+\frac{4}{n}}{6+\frac{7}{n}}=\frac{2+0}{6+0}=\frac{1}{3}$  

Zatem mamy 

$\underset{n\to \infty }{\lim }\frac{\frac{1-4n}{2n}-\frac{n}{n+5}}{{\left(1-\frac{2n+4}{6n+7}\right)}^2}=\frac{\underset{n\to \infty }{\lim }\left(\frac{1-4n}{2n}-\frac{n}{n+5}\right)}{\underset{n\to \infty }{\lim }\left({\left(1-\frac{2n+4}{6n+7}\right)}^2\right)}=\frac{\underset{n\to \infty }{\lim }\frac{1-4n}{2n}-\underset{n\to \infty }{\lim }\frac{n}{n+5}}{{\left(\underset{n\to \infty }{\lim }\left(1-\frac{2n+4}{6n+7}\right)\right)}^2}=$  

$=\frac{\underset{n\to \infty }{\lim }\frac{1-4n}{2n}-\underset{n\to \infty }{\lim }\frac{n}{n+5}}{{\left(1-\underset{n\to \infty }{\lim }\frac{2n+4}{6n+7}\right)}^2}=\frac{-2-1}{{\left(1-\frac{1}{3}\right)}^2}=\frac{-3}{{\left(\frac{2}{3}\right)}^2}=\frac{-3}{\frac{4}{9}}=-\frac{27}{4}=-6\frac{3}{4}$