$a\text{)}\underset{n\to \infty }{\lim }\left(\frac{n+3}{4}-\frac{n+5}{3}\right)=\underset{n\to \infty }{\lim }\left(\frac{3n+9}{12}-\frac{4n+20}{12}\right)=$  

$=\underset{n\to \infty }{\lim }\left(\frac{3n+9}{12}-\frac{4n+20}{12}\right)=\underset{n\to \infty }{\lim }\left(\frac{3n+9-4n-20}{12}\right)=$  

$=\underset{n\to \infty }{\lim }\left(\frac{-n-11}{12}\right)=\underset{n\to \infty }{\lim }\left(n\cdot \frac{-1-\frac{11}{n}}{12}\right)=$  

$=\underset{n\to \infty }{\lim }n\cdot \underset{n\to \infty }{\lim }\frac{-1-\frac{11}{n}}{12}$  

Zauważmy, że:

$\underset{n\to \infty }{\lim }n=+\infty$  

$\underset{n\to \infty }{\lim }\frac{-1-\frac{11}{n}}{12}=-\frac{1}{12}$  

Na mocy twierdzenia 3 zamieszczonego w podręczniku na stronie 153 otrzymujemy, że:

$=\underset{n\to \infty }{\lim }n\cdot \underset{n\to \infty }{\lim }\frac{-1-\frac{11}{n}}{12}=-\infty$  

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

$=\underset{n\to \infty }{\lim }\frac{4n^3-4n^3-6}{2n}=\underset{n\to \infty }{\lim }\frac{-6}{2n}=\underset{n\to \infty }{\lim }\frac{-3}{n}=0$  

 

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$c\text{)}\underset{n\to \infty }{\lim }\left(\frac{8}{n}+\frac{4n+3n^3}{n^2}\right)=\underset{n\to \infty }{\lim }\left(\frac{8n}{n^2}+\frac{4n+3n^3}{n^2}\right)=$  

$=\underset{n\to \infty }{\lim }\frac{8n+4n+3n^3}{n^2}=\underset{n\to \infty }{\lim }\frac{3n^3+12n}{n^2}=$  

$=\underset{n\to \infty }{\lim }\frac{\frac{3n^3}{n^2}+\frac{12n}{n^2}}{\frac{n^2}{n^2}}=\underset{n\to \infty }{\lim }\frac{3n+\frac{12}{n}}{1}=$  

$=\underset{n\to \infty }{\lim }\left(3n+\frac{12}{n}\right)=\underset{n\to \infty }{\lim }\left(3n\right)+\underset{n\to \infty }{\lim }\frac{12}{n}=+\infty$  

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

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

$=\underset{n\to \infty }{\lim }\frac{5n^2-2n-1}{n^2}=\underset{n\to \infty }{\lim }\frac{\frac{5n^2}{n^2}-\frac{2n}{n^2}-\frac{1}{n^2}}{\frac{n^2}{n^2}}=\underset{n\to \infty }{\lim }\frac{5-\frac{2}{n}-\frac{1}{n^2}}{1}=$  

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

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$e\text{)}\underset{n\to \infty }{\lim }\left[\frac{n-1}{2}-\frac{1+3+5+\dots +\left(2n-1\right)}{n+1}\right]$

Zauważmy, że liczby:

$1,3,5,\dots ,\left(2n-1\right)$   

Tworzą ciąg arytmetyczny taki, że:

$\{\begin{array}{l}{a}_1=1\\ r=2\end{array}$  

Zatem:

$1+3+5+\dots +\left(2n-1\right)=\frac{2\cdot 1+\left(n-1\right)\cdot 2}{2}\cdot n$  

Czyli:

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

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

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

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

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

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

Zauważmy, że:

$\underset{n\to \infty }{\lim }n=+\infty$  

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

Na mocy twierdzenia 3 zamieszczonego w podręczniku na stronie 153 otrzymujemy, że:

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

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$f\text{)}\underset{n\to \infty }{\lim }\frac{4+7+10+\dots +\left(3n+1\right)}{2n}$

Zauważmy, że liczby:

$4,7,10,\dots ,\left(3n+1\right)$   

Tworzą ciąg arytmetyczny taki, że:

$\{\begin{array}{l}{a}_1=4\\ r=3\end{array}$  

Zatem:

$4+7+10+\dots +\left(3n+1\right)=\frac{2\cdot 4+\left(n-1\right)\cdot 3}{2}\cdot n$  

Czyli:

$\underset{n\to \infty }{\lim }\frac{4+7+10+\dots +\left(3n+1\right)}{2n}=\underset{n\to \infty }{\lim }\left[\frac{\frac{2\cdot 4+\left(n-1\right)\cdot 3}{2}\cdot n}{2n}\right]=$  

$=\underset{n\to \infty }{\lim }\frac{\left(8+\left(n-1\right)\cdot 3\right)\cdot n}{4n}=\underset{n\to \infty }{\lim }\frac{\left(8+3n-3\right)}{4}=\underset{n\to \infty }{\lim }\frac{\left(5+3n\right)}{4}=$  

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

 

Zauważmy, że:

$\underset{n\to \infty }{\lim }n=+\infty$  

$\underset{n\to \infty }{\lim }\frac{\frac{5}{n}+3}{4}=\frac{3}{4}$  

Na mocy twierdzenia 3 zamieszczonego w podręczniku na stronie 153 otrzymujemy, że:

$=\underset{n\to \infty }{\lim }n\cdot \underset{n\to \infty }{\lim }\frac{\frac{5}{n}+3}{4}=+\infty$