Metoda 1)(Dam factor comun fortat la numarator)
[tex] \displaystyle \lim_{x \to \infty} \frac{x^2(1+ \frac{2}{x}+ \frac{1}{x^2} ) }{x} =\lim_{x \to \infty} x(1+ \frac{2}{x}+ \frac{1}{x^2} )=+\infty[/tex]
Metoda 2)(Aplicam teorema lui l'Hospital)
[tex]\displaystyle \lim_{x \to \infty} \frac{x^2+2x+1}{x}= \frac{+\infty}{+\infty} =\lim_{x \to \infty} \frac{(x^2+2x+1)'}{x'}=\\= \lim_{x \to \infty} \frac{2x+2}{1}= \lim_{x \to \infty} (2x+2)=+\infty[/tex]
