Przypomnijmy:

Prosta o równaniu y=ax+b jest asymptotą ukośną prawostronną (odpowiednio lewostronną) wykresu funkcji f wtedy i tylko wtedy, gdy:

$\underset{x\to \pm \infty }{\lim }\frac{f\left(x\right)}{x}=a\wedge \underset{x\to \pm \infty }{\lim }\left[f\left(x\right)-ax\right]=b.$  

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a)

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

$D_f=\mathbb{R}$  

 

Wyznaczmy asymptotę lewostronną:

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

 

$\underset{x\to -\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to -\infty }{\lim }\left[\frac{3x^2+1}{\left|x\right|+5}+3x\right]=\underset{x\to -\infty }{\lim }\left[\frac{3x^2+1}{-x+5}+\frac{3x\left(-x+5\right)}{-x+5}\right]=\underset{x\to -\infty }{\lim }\left[\frac{3x^2+1}{-x-5}+\frac{-3x^2+15x}{-x+5}\right]=\underset{x\to -\infty }{\lim }\frac{3x^2+1-3x^2+15x}{-x+5}=\underset{x\to -\infty }{\lim }\frac{15x+1}{-x+5}=\underset{x\to -\infty }{\lim }\frac{x\left(15+\frac{1}{x}\right)}{x\left(-1+\frac{5}{x}\right)}=\underset{x\to -\infty }{\lim }\frac{15+\frac{1}{x}}{-1+\frac{5}{x}}=\frac{15}{-1}=-15$  

 

$\underline{\underline{y=-3x-15}}$  

 

Wyznaczmy asymptotę prawostronną:

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

 

$\underset{x\to +\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to +\infty }{\lim }\left[\frac{3x^2+1}{\left|x\right|+5}-3x\right]=\underset{x\to +\infty }{\lim }\left[\frac{3x^2+1}{x+5}-\frac{3x\left(x+5\right)}{x+5}\right]=\underset{x\to +\infty }{\lim }\left[\frac{3x^2+1}{x-5}-\frac{3x^2+15x}{x+5}\right]=\underset{x\to +\infty }{\lim }\frac{3x^2+1-3x^2-15x}{x+5}=\underset{x\to +\infty }{\lim }\frac{-15x+1}{x+5}=\underset{x\to +\infty }{\lim }\frac{x\left(-15+\frac{1}{x}\right)}{x\left(1+\frac{5}{x}\right)}=\underset{x\to +\infty }{\lim }\frac{-15+\frac{1}{x}}{1+\frac{5}{x}}=\frac{-15}{1}=-15$  

 

$\underline{\underline{y=3x-15}}$  

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b)

$f\left(x\right)=\frac{4x-3}{\left|x\right|+2}$  

$D_f=\mathbb{R}$  

 

Wyznaczmy asymptotę lewostronną:

$\underset{x\to -\infty }{\lim }\frac{f\left(x\right)}{x}=\underset{x\to -\infty }{\lim }\frac{\frac{4x-3}{|x|+2}}{x}=\underset{x\to -\infty }{\lim }\frac{4x-3}{x\left|x\right|+2x}=\underset{x\to -\infty }{\lim }\frac{4x-3}{-x^2+2x}=\underset{x\to -\infty }{\lim }\frac{x^2\left(\frac{4}{x}-\frac{3}{x^2}\right)}{x^2\left(-1+\frac{2}{x}\right)}=\underset{x\to -\infty }{\lim }\frac{\frac{4}{x}-\frac{3}{x^2}}{-1+\frac{2}{x}}=\frac{0}{-1}=0$  

 

$\underset{x\to -\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to -\infty }{\lim }\left[\frac{4x-3}{\left|x\right|+2}-0\right]=\underset{x\to -\infty }{\lim }\frac{4x-3}{-x+2}=\underset{x\to -\infty }{\lim }\frac{x\left(4-\frac{3}{x}\right)}{x\left(-1+\frac{2}{x}\right)}=\underset{x\to -\infty }{\lim }\frac{4-\frac{3}{x}}{-1+\frac{2}{x}}=\frac{4}{-1}=-4$  

 

$\underline{\underline{y=-4}}$

 

Wyznaczmy asymptotę prawostronną:

$\underset{x\to +\infty }{\lim }\frac{f\left(x\right)}{x}=\underset{x\to +\infty }{\lim }\frac{\frac{4x-3}{|x|+2}}{x}=\underset{x\to +\infty }{\lim }\frac{4x-3}{x\left|x\right|+2x}=\underset{x\to +\infty }{\lim }\frac{4x-3}{x^2+2x}=\underset{x\to +\infty }{\lim }\frac{x^2\left(\frac{4}{x}-\frac{3}{x^2}\right)}{x^2\left(1+\frac{2}{x}\right)}=\underset{x\to +\infty }{\lim }\frac{\frac{4}{x}-\frac{3}{x^2}}{1+\frac{2}{x}}=\frac{0}{1}=0$  

 

$\underset{x\to +\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to +\infty }{\lim }\left[\frac{4x-3}{\left|x\right|+2}-0\right]=\underset{x\to +\infty }{\lim }\frac{4x-3}{x+2}=\underset{x\to +\infty }{\lim }\frac{x\left(4-\frac{3}{x}\right)}{x\left(1+\frac{2}{x}\right)}=\underset{x\to +\infty }{\lim }\frac{4-\frac{3}{x}}{1+\frac{2}{x}}=\frac{4}{1}=4$  

 

$\underline{\underline{y=4}}$  

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c)

$f\left(x\right)=\frac{16-x^2}{\left|x+2\right|-2}$  

Zał:

$\left|x+2\right|-2\ne 0$  

$\left|x+2\right|\ne 2$  

$x+2\ne 2,x+2\ne -2$  

$x\ne 0,x\ne -4$  

$D_f=\mathbb{R}-\left\{-4,0\right\}$  

 

Wyznaczmy asymptotę lewostronną:

$\underset{x\to -\infty }{\lim }\frac{f\left(x\right)}{x}=\underset{x\to -\infty }{\lim }\frac{\frac{16-x^2}{|x+2|-2}}{x}=\underset{x\to -\infty }{\lim }\frac{16-x^2}{x\left|x+2\right|-2x}=\underset{x\to -\infty }{\lim }\frac{16-x^2}{-x^2-2x-2x}=\underset{x\to -\infty }{\lim }\frac{16-x^2}{-x^2-4x}=\underset{x\to -\infty }{\lim }\frac{x^2\left(\frac{16}{x^2}-1\right)}{x^2\left(-1-\frac{4}{x}\right)}=\underset{x\to -\infty }{\lim }\frac{\frac{16}{x^2}-1}{-1-\frac{4}{x}}=\frac{-1}{-1}=1$  

 

$\underset{x\to -\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to -\infty }{\lim }\left[\frac{16-x^2}{\left|x+2\right|-2}-x\right]=\underset{x\to -\infty }{\lim }\left[\frac{16-x^2}{-x-2-2}-x\right]=\underset{x\to -\infty }{\lim }\left[\frac{16-x^2}{-x-4}-\frac{x\left(-x-4\right)}{-x-4}\right]=\underset{x\to -\infty }{\lim }\left[\frac{16-x^2+x^2+4x}{-x-4}\right]=\underset{x\to -\infty }{\lim }\frac{16+4x}{-x-4}=\underset{x\to -\infty }{\lim }\frac{x\left(\frac{16}{x}+4\right)}{x\left(-1-\frac{4}{x}\right)}=\frac{4}{-1}=-4$  

 

$\underline{\underline{y=1x-4}}$  

 

Wyznaczmy asymptotę prawostronną:

$\underset{x\to +\infty }{\lim }\frac{f\left(x\right)}{x}=\underset{x\to +\infty }{\lim }\frac{\frac{16-x^2}{|x+2|-2}}{x}=\underset{x\to +\infty }{\lim }\frac{16-x^2}{x\left|x+2\right|-2x}=\underset{x\to +\infty }{\lim }\frac{16-x^2}{x^2+2x-2x}=\underset{x\to +\infty }{\lim }\frac{16-x^2}{x^2}=\underset{x\to +\infty }{\lim }\frac{x^2\left(\frac{16}{x^2}-1\right)}{x^2}=\underset{x\to +\infty }{\lim }\frac{\frac{16}{x^2}-1}{1}=\frac{-1}{1}=-1$

 

$\underset{x\to +\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to +\infty }{\lim }\left[\frac{16-x^2}{\left|x+2\right|-2}-\left(-x\right)\right]=\underset{x\to +\infty }{\lim }\left[\frac{16-x^2}{x+2-2}+x\right]=\underset{x\to +\infty }{\lim }\left[\frac{16-x^2}{x}+\frac{x^2}{x}\right]=\underset{x\to +\infty }{\lim }\left[\frac{16-x^2+x^2}{x}\right]=\underset{x\to +\infty }{\lim }\frac{16}{x}=\underset{x\to +\infty }{\lim }\left(16\cdot \frac{1}{x}\right)=0$  

 

$\underline{\underline{y=-1x}}$  

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d)

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

Zał:

$\left|x+1\right|+1\ne 0$  

$\left|x+1\right|\ne -1$  

$D_f=\mathbb{R}$  

 

Wyznaczmy asymptotę lewostronną:

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

 

$\underset{x\to -\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to -\infty }{\lim }\left[\frac{x^2-4}{\left|x+1\right|+1}+x\right]=\underset{x\to -\infty }{\lim }\left[\frac{x^2-4}{-x}+\frac{-x^2}{-x}\right]=\underset{x\to -\infty }{\lim }\left[\frac{x^2-4-x^2}{-x}\right]=\underset{x\to -\infty }{\lim }\frac{-4}{-x}=0$

 

$y=-1x$  

 

Wyznaczmy asymptotę prawostronną:

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

 

$\underset{x\to +\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to +\infty }{\lim }\left[\frac{x^2-4}{\left|x+1\right|+1}-x\right]=\underset{x\to +\infty }{\lim }\left[\frac{x^2-4}{x+2}-\frac{x\left(x+2\right)}{x+2}\right]=\underset{x\to +\infty }{\lim }\left[\frac{x^2-4}{x+2}-\frac{x^2+2x}{x+2}\right]=\underset{x\to +\infty }{\lim }\frac{x^2-4-x^2-2x}{x+2}=\underset{x\to +\infty }{\lim }\frac{-2x-4}{x+2}=\underset{x\to +\infty }{\lim }\frac{-2\left(x+2\right)}{x+2}=-2$  

 

$\underline{\underline{y=1x-2}}$  

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e)

$f\left(x\right)=\sqrt{x^2+6x}$  

Zał:

$x^2+6x\ge 0$  

$x\left(x+6\right)\ge 0$  

$x\in (-\infty ,-6⟩\cup ⟨0,+\infty )$  

$D_f=(-\infty ,-6⟩\cup ⟨0,+\infty )$  

 

Wyznaczmy asymptotę lewostronną:

$\underset{x\to -\infty }{\lim }\frac{f\left(x\right)}{x}=\underset{x\to -\infty }{\lim }\frac{\sqrt{x^2+6x}}{x}=\underset{x\to -\infty }{\lim }\frac{\sqrt{x^2\left(1+\frac{6}{x}\right)}}{x}=\underset{x\to -\infty }{\lim }\frac{\left|x\right|\sqrt{1+\frac{6}{x}}}{x}=\underset{x\to -\infty }{\lim }\frac{-x\sqrt{1+\frac{6}{x}}}{x}=\underset{x\to -\infty }{\lim }\left(-1\sqrt{1+\frac{6}{x}}\right)=-1$  

 

$\underset{x\to -\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to -\infty }{\lim }\left[\sqrt{x^2+6x}+x\right]=\underset{x\to -\infty }{\lim }\left[\left(\sqrt{x^2+6x}+x\right)\cdot \frac{\sqrt{x^2+6x}-x}{\sqrt{x^2+6x}-x}\right]=\underset{x\to -\infty }{\lim }\left[\frac{{\sqrt{x^2+6x}}^2-x^2}{\sqrt{x^2+6x}-x}\right]=\underset{x\to -\infty }{\lim }\left[\frac{x^2+6x-x^2}{\sqrt{x^2\left(1+\frac{6}{x}\right)}-x}\right]=\underset{x\to -\infty }{\lim }\frac{6x}{\left|x\right|\sqrt{1+\frac{6}{x}}-x}=\underset{x\to -\infty }{\lim }\frac{6x}{-x\sqrt{1+\frac{6}{x}}-x}=\underset{x\to -\infty }{\lim }\frac{6x}{x\left(-\sqrt{1+\frac{6}{x}}-1\right)}=\underset{x\to -\infty }{\lim }\frac{6}{-\sqrt{1+\frac{6}{x}}-1}=\frac{6}{-2}=-3$  

 

$\underline{\underline{y=-1x-3}}$  

 

Wyznaczmy asymptotę prawostronną:

$\underset{x\to +\infty }{\lim }\frac{f\left(x\right)}{x}=\underset{x\to +\infty }{\lim }\frac{\sqrt{x^2+6x}}{x}=\underset{x\to +\infty }{\lim }\frac{\sqrt{x^2\left(1+\frac{6}{x}\right)}}{x}=\underset{x\to +\infty }{\lim }\frac{\left|x\right|\sqrt{1+\frac{6}{x}}}{x}=\underset{x\to +\infty }{\lim }\frac{x\sqrt{1+\frac{6}{x}}}{x}=\underset{x\to +\infty }{\lim }\sqrt{1+\frac{6}{x}}=1$  

 

$\underset{x\to +\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to +\infty }{\lim }\left[\sqrt{x^2+6x}-x\right]=\underset{x\to +\infty }{\lim }\left[\left(\sqrt{x^2+6x}-x\right)\cdot \frac{\sqrt{x^2+6x}+x}{\sqrt{x^2+6x}+x}\right]=\underset{x\to +\infty }{\lim }\left[\frac{{\sqrt{x^2+6x}}^2-x^2}{\sqrt{x^2+6x}+x}\right]=\underset{x\to +\infty }{\lim }\left[\frac{x^2+6x-x^2}{\sqrt{x^2\left(1+\frac{6}{x}\right)}+x}\right]=\underset{x\to +\infty }{\lim }\frac{6x}{\left|x\right|\sqrt{1+\frac{6}{x}}+x}=\underset{x\to +\infty }{\lim }\frac{6x}{x\sqrt{1+\frac{6}{x}}+x}=\underset{x\to +\infty }{\lim }\frac{6x}{x\left(\sqrt{1+\frac{6}{x}}+1\right)}=\underset{x\to +\infty }{\lim }\frac{6}{\sqrt{1+\frac{6}{x}}+1}=\frac{6}{2}=3$  

 

$\underline{\underline{y=1x+3}}$  

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f)

$f\left(x\right)=\sqrt{4x^2-x}$  

Zał:

$4x^2-x\ge 0$  

$x\left(4x-1\right)\ge 0$  

$x\in (-\infty ,0⟩\cup ⟨\frac{1}{4},+\infty )$  

$D_f=(-\infty ,0⟩\cup ⟨\frac{1}{4},+\infty )$  

 

Wyznaczmy asymptotę lewostronną:

$\underset{x\to -\infty }{\lim }\frac{f\left(x\right)}{x}=\underset{x\to -\infty }{\lim }\frac{\sqrt{4x^2-x}}{x}=\underset{x\to -\infty }{\lim }\frac{\sqrt{x^2\left(4-\frac{1}{x}\right)}}{x}=\underset{x\to -\infty }{\lim }\frac{\left|x\right|\sqrt{4-\frac{1}{x}}}{x}=\underset{x\to -\infty }{\lim }\frac{-x\sqrt{4-\frac{1}{x}}}{x}=\underset{x\to -\infty }{\lim }\left(-\sqrt{4-\frac{1}{x}}\right)=-\sqrt{4}=-2$  

 

$\underset{x\to -\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to -\infty }{\lim }\left[\sqrt{4x^2-x}+2x\right]=\underset{x\to -\infty }{\lim }\left[\left(\sqrt{4x^2-x}+2x\right)\cdot \frac{\sqrt{4x^2-x}-2x}{\sqrt{4x^2-x}-2x}\right]=\underset{x\to -\infty }{\lim }\left[\frac{{\sqrt{4x^2-x}}^2-{\left(2x\right)}^2}{\sqrt{x^2\left(4-\frac{1}{x}\right)}-2x}\right]=\underset{x\to -\infty }{\lim }\frac{4x^2-x-4x^2}{\left|x\right|\sqrt{4-\frac{1}{x}}-2x}=\underset{x\to -\infty }{\lim }\frac{-x}{-x\sqrt{4-\frac{1}{x}}-2x}=\underset{x\to -\infty }{\lim }\frac{-x}{-x\left(\sqrt{4-\frac{1}{x}}+2\right)}=\underset{x\to -\infty }{\lim }\frac{1}{\sqrt{4-\frac{1}{x}}+2}=\frac{1}{\sqrt{4}+2}=\frac{1}{4}$  

 

$\underline{\underline{y=-2x+\frac{1}{4}}}$  

 

Wyznaczmy asymptotę prawostronną:

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

 

$\underset{x\to +\infty }{\lim }\left[f\left(x\right)-ax\right]=\underset{x\to +\infty }{\lim }\left[\sqrt{4x^2-x}-2x\right]=\underset{x\to +\infty }{\lim }\left[\left(\sqrt{4x^2-x}-2x\right)\cdot \frac{\sqrt{4x^2-x}+2x}{\sqrt{4x^2-x}+2x}\right]=\underset{x\to +\infty }{\lim }\left[\frac{{\sqrt{4x^2-x}}^2-{\left(2x\right)}^2}{\sqrt{x^2\left(4-\frac{1}{x}\right)}+2x}\right]=\underset{x\to +\infty }{\lim }\frac{4x^2-x-4x^2}{\left|x\right|\sqrt{4-\frac{1}{x}}+2x}=\underset{x\to +\infty }{\lim }\frac{-x}{x\sqrt{4-\frac{1}{x}}+2x}=\underset{x\to +\infty }{\lim }\frac{-x}{x\left(\sqrt{4-\frac{1}{x}}+2\right)}=\underset{x\to +\infty }{\lim }\frac{-1}{\sqrt{4-\frac{1}{x}}+2}=\frac{-1}{\sqrt{4}+2}=\frac{-1}{4}$  

 

$\underline{\underline{y=2x-\frac{1}{4}}}$