Odpowiedź:
a ) f ( x ) = [tex]\frac{2 x + 1}{x - 1} = \frac{2*(x - 1) + 3}{x - 1} = 2 + \frac{3}{x - 1}[/tex]
więc
[tex]\lim_{x \to \infty} f( x )= 2[/tex]
[tex]\lim_{x \to - \infty} f ( x ) = 2[/tex]
[tex]\lim_{x \to 1^{-}} f ( x ) =[/tex] - ∞
[tex]\lim_{x \to 1^{+}} f ( x ) = +[/tex] ∞
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Odpowiedź:
a ) f ( x ) = [tex]\frac{2 x + 1}{x - 1} = \frac{2*(x - 1) + 3}{x - 1} = 2 + \frac{3}{x - 1}[/tex]
więc
[tex]\lim_{x \to \infty} f( x )= 2[/tex]
[tex]\lim_{x \to - \infty} f ( x ) = 2[/tex]
[tex]\lim_{x \to 1^{-}} f ( x ) =[/tex] - ∞
[tex]\lim_{x \to 1^{+}} f ( x ) = +[/tex] ∞
======================
Szczegółowe wyjaśnienie: