Răspuns:
Explicație pas cu pas:
x=x0 este asimptota vericala daca x -> x0-0 si x->x0+0, f(x)->±∞
D=R\{-1;1}[tex]f(x)=\frac{x}{x^{2}-1}=\frac{x}{(x-1)(x+1)}\\ \lim_{x \to -1-0} \frac{x}{(x-1)(x+1)}=\frac{-1}{-2*(-0)} =-\infty\\ \lim_{x \to -1+0} \frac{x}{(x-1)(x+1)}=\frac{-1}{-2*(+0)} =+\infty\\~deci~x=-1~este~asimptota~verticala\\ \lim_{x \to 1-0} \frac{x}{(x-1)(x+1)}=\frac{1}{(-0)*2} =-\infty\\ \lim_{x \to 1+0} \frac{x}{(x-1)(x+1)}=\frac{1}{(+0)*2} =+\infty\\~deci~x=1~este~asimptota~verticala\\[/tex]
