Question

Fie f:R/(1)-R , f(x)=x^2 / x-1 .Aflati asimptotele la Gf .

Answer 1

[tex]\displaystyle \lim_{x \to\pm \infty} \frac{x^2}{x-1}=\pm \infty=\ \textgreater \ f\ nu\ admite\ as.\ orizontale\\ \lim_{x \to 1,x\ \textless \ 1} \frac{x^2}{x-1}=-\infty=\ \textgreater \ x=1\ as. verticala\ la\ -\infty\\ \lim_{x \to 1,x\ \textgreater \ 1} \frac{x^2}{x-1}=+\infty=\ \textgreater \ x=1\ as. verticala\ la\ +\infty\\ y=mx+n\\ m=\lim_{x \to \pm \infty} \frac{f(x)}{x}=\lim_{x \to \pm \infty} \frac{x^2}{x^2-x}=1\\ n=\lim_{x \to \pm \infty}[f(x)-mx]=\lim_{x \to \pm \infty}(\frac{x^2}{x-1} -x)=\\ =\lim_{x \to \pm \infty}\frac{x}{x-1} =1\\ y=x+1\ as.oblica[/tex]