Grenseverdier for en brøk når den variable går mot uendelig

$\underset{x\to \pm \infty }{\lim }\frac{x}{x+1}=\underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{x}{x}}}{{\displaystyle \frac{x}{x}}+{\displaystyle \frac{1}{x}}}=\underset{x\to \pm \infty }{\lim }\frac{{\displaystyle 1}}{{\displaystyle 1+\frac{1}{x}}}=\frac{1}{1+0}=1$

$\underset{x\to \pm \infty }{\lim }\frac{2}{x-1}=\underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{2}{x}}}{{\displaystyle \frac{x}{x}}-{\displaystyle \frac{1}{x}}}=\underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{2}{x}}}{{\displaystyle 1-\frac{1}{x}}}=\frac{0}{1-0}=0$

$\underset{x\to \pm \infty }{\lim }\frac{2x-3}{-5x+2}=\underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{2x}{x}}-{\displaystyle \frac{3}{x}}}{-{\displaystyle \frac{5x}{x}}+{\displaystyle \frac{2}{x}}}=\underset{x\to \pm \infty }{\lim }\frac{{\displaystyle 2-\frac{3}{x}}}{-{\displaystyle 5}+{\displaystyle \frac{2}{x}}}=\frac{2-0}{-5+0}=-\frac{2}{5}$

$\underset{x\to \pm \infty }{\lim }\frac{3x^2}{x+4}=\underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{3x^2}{x^2}}}{{\displaystyle \frac{x}{x^2}}+{\displaystyle \frac{4}{x^2}}}=\underset{x\to \pm \infty }{\lim }\frac{{\displaystyle 3}}{{\displaystyle \frac{1}{x}+\frac{4}{x^2}}}=\frac{3}{0+0}$

Grenseverdien eksisterer ikke.

$\begin{array}{rcl}\underset{x\to \pm \infty }{\lim }\frac{3x-6}{x^2-5x+6} & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{3x}{x^2}}-{\displaystyle \frac{6}{x^2}}}{{\displaystyle \frac{x^2}{x^2}}-{\displaystyle \frac{5x}{x^2}}+{\displaystyle \frac{6}{x^2}}}\\ & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{3}{x}}-{\displaystyle \frac{6}{x^2}}}{1-{\displaystyle \frac{5}{x}}+{\displaystyle \frac{6}{x^2}}}=\frac{0-0}{1}=0\end{array}$

Løsning med CAS i GeoGebra:

$\begin{array}{rcl}\underset{x\to \pm \infty }{\lim }\frac{3x^2-6}{x^2-5x+6} & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{{\displaystyle 3x^2}}{x^2}-\frac{{\displaystyle 6}}{x^2}}}{{\displaystyle \frac{{\displaystyle x^2}}{x^2}-\frac{{\displaystyle 5x}}{x^2}+\frac{{\displaystyle 6}}{x^2}}}\\ & =& \underset{x\to \pm \infty }{\lim }\frac{3-{\displaystyle \frac{6}{x^2}}}{{\displaystyle 1-\frac{5}{x}+\frac{6}{x^2}}}=\frac{3}{1}=3\end{array}$

$\begin{array}{rcl}\underset{x\to \pm \infty }{\lim }\frac{3x^2-6}{x^3-5x+6} & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{3x^2}{x^3}}-{\displaystyle \frac{6}{x^3}}}{{\displaystyle \frac{x^3}{x^3}}-{\displaystyle \frac{5x}{x^3}}+{\displaystyle \frac{6}{x^3}}}\\ & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{3}{x}}-{\displaystyle \frac{6}{x^3}}}{1-{\displaystyle \frac{5}{x^2}}+{\displaystyle \frac{6}{x^3}}}=\frac{0-0}{1-0+0}=0\end{array}$

$\begin{array}{rcl}\underset{x\to \pm \infty }{\lim }\frac{3x^3-6}{x^3-5x+6} & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{3x^3}{x^3}-\frac{{\displaystyle 6}}{x^3}}}{{\displaystyle \frac{x^3}{x^3}-\frac{5x}{x^3}+\frac{{\displaystyle 6}}{x^3}}}\\ & =& \underset{x\to \pm \infty }{\lim }\frac{3-{\displaystyle \frac{6}{x^3}}}{1-{\displaystyle \frac{5}{x^2}}+{\displaystyle \frac{6}{x^3}}}=\frac{3-0}{1-0+0}=3\end{array}$

$\begin{array}{rcl}\underset{x\to \pm \infty }{\lim }\frac{8-4x^2}{8x^2+3x} & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{8}{x^2}}-{\displaystyle \frac{4x^2}{x^2}}}{{\displaystyle \frac{8x^2}{x^2}}+{\displaystyle \frac{3x}{x^2}}}\\ & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{8}{x^2}}-4}{8+{\displaystyle \frac{3}{x}}}=\frac{0-4}{8+0}=-\frac{1}{2}\end{array}$

$\begin{array}{rcl}\underset{x\to \pm \infty }{\lim }\frac{x^3-2x+5}{3-6x^3} & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle \frac{{\displaystyle x^3}}{x^3}-\frac{{\displaystyle 2x}}{x^3}+\frac{{\displaystyle 5}}{x^3}}}{{\displaystyle \frac{{\displaystyle 3}}{x^3}-\frac{{\displaystyle 6x^3}}{x^3}}}\\ & =& \underset{x\to \pm \infty }{\lim }\frac{{\displaystyle 1-\frac{2}{x^2}+\frac{5}{x^3}}}{{\displaystyle \frac{3}{x^3}-6}}=\frac{1-0+0}{0-6}=-\frac{1}{6}\end{array}$