Răspuns:
1)
Nu are asimptota orizontala ,deoarece functia tinde spre infinit.
Astfel, asimptota oblica:
[tex]y = mx + n[/tex]
[tex]m = \lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \frac{x^2+1}{x^2-x} = 1[/tex]
[tex]n = \lim_{x \to \infty} (f(x) - mx) = \lim_{x \to \infty} (\frac{x^2+1}{x-1} - x) = \lim_{x \to \infty} (\frac{x^2+1}{x-1} - x\frac{x-1}{x-1} ) = \lim_{x \to \infty} (\frac{x^2+1-x^2+x}{x-1} ) = \lim_{x \to \infty} (\frac{1+x}{x-1} ) =1[/tex]
y = x + 1 e asimptota oblica spre ±∞
Asimptota verticala:
x - 1 ≠ 0 => x ≠ 1 , deci functia e definita in felul urmator:
f: R \ {1} -> R , 1 fiind punct de acumulare.
Calculam limitele laterale in 1.
[tex]\lim_{x_{x>1} \to \ 1} f(x) = \lim_{x_{x>1} \to \ 1} \frac{x^2+1}{x-1} = \frac{2}{0+} = +infinit\\ \lim_{x_{x<1} \to \ 1} f(x) = \lim_{x_{x<1} \to \ 1} \frac{x^2+1}{x-1} = \frac{2}{0-} = -infinit\\[/tex]
Deci x = 1 e asimptota verticala.
2)
Asimptota orizontala spre +∞
[tex]\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x-1}{\sqrt{x^2+4} } = \lim_{x \to \infty} \frac{x-1}{|x|\sqrt{1+\frac{4}{x^2} } } = \lim_{x \to \infty} \frac{x(1-\frac{1}{x} )}{x\sqrt{1+\frac{4}{x^2} } } = 1[/tex]
y = 1
Asimpt. orizontala spre -∞
[tex]\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x-1}{\sqrt{x^2+4} } = \lim_{x \to \infty} \frac{x-1}{|x|\sqrt{1+\frac{4}{x^2} } } = \lim_{x \to \infty} \frac{x(1-\frac{1}{x} )}{-x\sqrt{1+\frac{4}{x^2} } } = -1[/tex]
y = -1
Nu are asimptota oblica, nici verticala. ( nu exista puncte de acumulare).
