Răspuns:
Explicație pas cu pas:
[tex]\displaystyle\limit\lim_{x\to\infty}\dfrac{2x^2}{3x+x^2+1}=\lim_{x\to\infty}\dfrac{2x^2}{x^2\left(\dfrac{3}{x}+1+\dfrac{1}{x^2}\right)}=\boxed{2}\\\texttt{De remarcat faptul ca }\lim_{x\to\infty}\dfrac{3}{x}=\lim_{x\to\infty}\dfrac{1}{x^2}=0[/tex]
