Odpowiedź:
f ( x) = [tex]\frac{x^2 - 1}{( 2 x - 1)^2} = \frac{x^2 - 1}{4 x^2 - 4 x +1}[/tex] x ≠ [tex]\frac{1}{2}[/tex]
Mamy
[tex]\lim_{x \to \frac{1}{2}^- } f(x) =[/tex] [tex]\frac{-3/4}{0^+} = -oo[/tex]
[tex]\lim_{x \to \frac{1}{2}^+ } f ( x) = \frac{-3/4}{0^+} = - oo[/tex]
Asymptota pionowa :
x = [tex]\frac{1}{2}[/tex]
===================
f ( x) = [tex]\frac{1 - \frac{1}{x^{2} } }{4 - \frac{4}{x} + \frac{1}{x^{2} } }[/tex]
więc
[tex]\lim_{x \to - \infty} f( x) = \frac{1 - 0}{4 - 0 + 0} = \frac{1}{4}[/tex]
[tex]\lim_{n \to+ \infty} f ( x) = \frac{1}{4}[/tex]
Asymptota pozioma:
y = [tex]\frac{1}{4}[/tex]
====================
Szczegółowe wyjaśnienie:
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Odpowiedź:
f ( x) = [tex]\frac{x^2 - 1}{( 2 x - 1)^2} = \frac{x^2 - 1}{4 x^2 - 4 x +1}[/tex] x ≠ [tex]\frac{1}{2}[/tex]
Mamy
[tex]\lim_{x \to \frac{1}{2}^- } f(x) =[/tex] [tex]\frac{-3/4}{0^+} = -oo[/tex]
[tex]\lim_{x \to \frac{1}{2}^+ } f ( x) = \frac{-3/4}{0^+} = - oo[/tex]
Asymptota pionowa :
x = [tex]\frac{1}{2}[/tex]
===================
f ( x) = [tex]\frac{1 - \frac{1}{x^{2} } }{4 - \frac{4}{x} + \frac{1}{x^{2} } }[/tex]
więc
[tex]\lim_{x \to - \infty} f( x) = \frac{1 - 0}{4 - 0 + 0} = \frac{1}{4}[/tex]
[tex]\lim_{n \to+ \infty} f ( x) = \frac{1}{4}[/tex]
Asymptota pozioma:
y = [tex]\frac{1}{4}[/tex]
====================
Szczegółowe wyjaśnienie: