Twierdzenie Niech funkcje f i g będą określone w sąsiedztwie punktu x0 oraz                                                                                                          $\underset{x\to x_0}{\lim }f\left(x\right)=a,a\in \mathbb{R},\underset{x\to x_0}{\lim }g\left(x\right)=+\infty$   Wówczas: $\underset{x\to x_0}{\lim }\left[f\left(x\right)+g\left(x\right)\right]=+\infty$   $\underset{x\to x_0}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=-\infty$     $\text{jesˊli}a>0,\text{to}\underset{x\to x_0}{\lim }\left[f\left(x\right)\cdot g\left(x\right)\right]=+\infty$    $\text{jesˊli}a<0,\text{to}\underset{x\to x_0}{\lim }\left[f\left(x\right)\cdot g\left(x\right)\right]=-\infty$

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a)

Obliczymy granicę 

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

z każdego z nawiasów wyłączamy x w najwyższej potędze i mamy 

${\left(1-x\right)}^2\left(x+3\right)=\left(1-2x+x^2\right)\left(x+3\right)=x^2\left(\frac{1}{x^2}-\frac{2}{x}+1\right)\cdot x\left(1+\frac{3}{x}\right)=x^3\left(\frac{1}{x^2}-\frac{2}{x}+1\right)\left(1+\frac{3}{x}\right)$  

Zauważmy, że 

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

Zatem mamy 

$\underset{x\to +\infty }{\lim }\left[\left(1-x^2\right)\left(x+3\right)\right]=\underset{x\to +\infty }{\lim }\left[\stackrel{\stackrel{+\infty }{\uparrow }}{x^3}\cdot \stackrel{\underset{\uparrow }{1}}{\overbrace{\left(\frac{1}{x^2}-\frac{2}{x}+1\right)\left(1+\frac{3}{x}\right)}}\right]=+\infty$  

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b)

Obliczymy granicę 

$\underset{x\to -\infty }{\lim }\left[-3x\left(2-x\right)\left(2+x\right)\right]$  

Korzystając ze wzoru skróconego mnożenia na różnicę kwadratów mamy 

$-3x\left(2-x\right)\left(2+x\right)=-3x\left(4-x^2\right)$  

wyłączamy z nawiasu x w najwyższej potędze i mamy 

$-3x\left(4-x^2\right)=-3x\cdot x^2\left(\frac{4}{x^2}-1\right)=-3x^3\left(\frac{4}{x^2}-1\right)$  

Zauważmy, że 

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

Zatem mamy 

$\underset{x\to -\infty }{\lim }\left[-3x\left(2-x\right)\left(2+x\right)\right]=\underset{x\to -\infty }{\lim }\left[\stackrel{\stackrel{+\infty }{\uparrow }}{-3x^3}\cdot \stackrel{\underset{\uparrow }{-1}}{\overbrace{\left(\frac{4}{x^2}-1\right)}}\right]=-\infty$  

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c)

Obliczymy granicę 

$\underset{x\to +\infty }{\lim }\left[{\left(x-3\right)}^3{\left(2-3x\right)}^2\right]$  

wyłączamy z nawiasu x w najwyższej potędze i mamy 

${\left(x-3\right)}^3{\left(2-3x\right)}^2={\left[x\left(1-\frac{3}{x}\right)\right]}^3\cdot {\left[x\left(\frac{2}{x}-3\right)\right]}^2=x^3{\left(1-\frac{3}{x}\right)}^3\cdot x^2{\left(\frac{2}{x}-3\right)}^2=x^5{\left(1-\frac{3}{x}\right)}^3{\left(\frac{2}{x}-3\right)}^2$  

Zauważmy, że 

• $\underset{x\to +\infty }{\lim }{x}^5=+\infty$  
• $\underset{x\to +\infty }{\lim }\left[{\left(1-\frac{3}{x}\right)}^3{\left(\frac{2}{x}-3\right)}^2\right]=1\cdot 9=9$  

Zatem mamy 

$\underset{x\to +\infty }{\lim }\left[{\left(x-3\right)}^3{\left(2-3x\right)}^2\right]=\underset{x\to +\infty }{\lim }\left[\stackrel{\stackrel{+\infty }{\uparrow }}{x^5}\cdot \stackrel{\underset{\uparrow }{9}}{\overbrace{{\left(1-\frac{3}{x}\right)}^3{\left(\frac{2}{x}-3\right)}^2}}\right]=+\infty$  

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d)

Obliczymy granicę 

$\underset{x\to -\infty }{\lim }\left[\left(5-x\right){\left(3-x\right)}^2\left(x+2\right)\right]$  

wyłączamy z nawiasu x w najwyższej potędze i mamy 

$\left(5-x\right){\left(3-x\right)}^2\left(x+2\right)=\left(5-x\right){\left[x\left(\frac{3}{x}-1\right)\right]}^2\left(x+2\right)=x\left(\frac{5}{x}-1\right)\cdot x^2{\left(\frac{3}{x}-1\right)}^2\cdot x\left(1+\frac{2}{x}\right)=$  

$=x^4\left(\frac{5}{x}-1\right){\left(\frac{3}{x}-1\right)}^2\left(1+\frac{2}{x}\right)$  

Zauważmy, że 

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

Zatem mamy 

$\underset{x\to -\infty }{\lim }\left[\left(5-x\right){\left(3-x\right)}^2\left(x+2\right)\right]=\underset{x\to -\infty }{\lim }\left[\stackrel{\stackrel{+\infty }{\uparrow }}{x^4}\cdot \stackrel{\underset{\uparrow }{-1}}{\overbrace{\left(\frac{5}{x}-1\right){\left(\frac{3}{x}-1\right)}^2\left(1+\frac{2}{x}\right)}}\right]=-\infty$  

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e)

Obliczymy granicę 

$\underset{x\to +\infty }{\lim }\left[\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\right]$  

Korzystając ze wzoru skróconego mnożenia na różnicę kwadratów mamy 

$\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)=1-x=x\left(\frac{1}{x}-1\right)$  

Zauważmy, że 

• $\underset{x\to +\infty }{\lim }x=+\infty$  
• $\underset{x\to +\infty }{\lim }\left(\frac{1}{x}-1\right)=-1$  

Zatem mamy 

$\underset{x\to +\infty }{\lim }\left[\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\right]=\underset{x\to +\infty }{\lim }\left[\stackrel{\stackrel{+\infty }{\uparrow }}{x}\cdot \stackrel{\underset{\uparrow }{-1}}{\overbrace{\left(\frac{1}{x}-1\right)}}\right]=-\infty$  

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f)

Obliczymy granicę 

$\underset{x\to -\infty }{\lim }\left[{\left(9-2x\right)}^3{\left(7-x\right)}^3\right]$  

Wyłączamy z każdego z nawiasów x w najwyższej potędze i otrzymujemy 

${\left(9-2x\right)}^3{\left(7-x\right)}^3={\left[x\left(\frac{9}{x}-2\right)\right]}^3{\left[x\left(\frac{7}{x}-1\right)\right]}^3=x^3\cdot {\left(\frac{9}{x}-2\right)}^3\cdot x^3\cdot {\left(\frac{7}{x}-1\right)}^3=x^6{\left(\frac{9}{x}-2\right)}^3{\left(\frac{7}{x}-1\right)}^3$  

Zauważmy, że 

• $\underset{x\to -\infty }{\lim }{x}^6=+\infty$  
• $\underset{x\to +\infty }{\lim }\left[{\left(\frac{9}{x}-2\right)}^3{\left(\frac{7}{x}-1\right)}^3\right]={\left(-2\right)}^3\cdot {\left(-1\right)}^3=8$  

Zatem mamy 

$\underset{x\to -\infty }{\lim }\left[{\left(9-2x\right)}^3{\left(7-x\right)}^3\right]=\underset{x\to -\infty }{\lim }\left[\stackrel{\stackrel{+\infty }{\uparrow }}{x^6}\cdot \stackrel{\underset{\uparrow }{8}}{\overbrace{{\left(\frac{9}{x}-2\right)}^3{\left(\frac{7}{x}-1\right)}^3}}\right]=+\infty$