Răspuns :
Cand x tinde la infinit, atunci fractiile cu x la numitor si numaratorul numar finit tind la 0. Rezolvarea este in poza.
[tex] \lim_{x \to \infty} \frac{2x^2-3x+5}{x^2+x+2}= \frac{\infty}{\infty} =ned.[/tex]
Aplicam Hospital
[tex] \lim_{x \to \infty} \frac{(2x^2-3x+5)`}{(x^2+x+2)`}= \lim_{x \to \infty} \frac{4x-3}{2x+1}= \frac{\infty}{\infty}=ned [/tex]
Aplicam l`hospital
[tex] \lim_{x \to \infty} \frac{(4x-3)`}{(2x+1)`}= \lim_{x \to \infty} \frac{4}{2}=2 [/tex]
Aplicam Hospital
[tex] \lim_{x \to \infty} \frac{(2x^2-3x+5)`}{(x^2+x+2)`}= \lim_{x \to \infty} \frac{4x-3}{2x+1}= \frac{\infty}{\infty}=ned [/tex]
Aplicam l`hospital
[tex] \lim_{x \to \infty} \frac{(4x-3)`}{(2x+1)`}= \lim_{x \to \infty} \frac{4}{2}=2 [/tex]