[tex]\lim_{x \to \pm\infty} \frac{2x^3+4x+6}{3x^3+2x+5} =\lim_{x \to \pm\infty} \frac{x^3(2+4/x^2+6/x^3)}{x^3(3+2/x^2+5/x^3)} = \\
=\lim_{x \to \pm\infty} \frac{(2+4/x^2+6/x^3)}{(3+2/x^2+5/x^3)} = \frac{2}{3} [/tex]
[tex] \lim_{x \to -1,x\ \textless \ -1 } \frac{2x^3+4x+6}{3x^3+2x+5}=\\ = \lim_{x \to -1,x\ \textless \ -1 \frac{2(x+1)(x^2-x+3)}{(x+1)(3x^2-3x+5)}= \\ \lim_{x \to -1,x\ \textless \ -1 \frac{2(x^2-x+3)}{3x^2-3x+5}= \frac{2(1+1+3)}{3+3+5}= \frac{10}{11}[/tex]
Analog pentru limita x->-1, x>-1.
