[tex]\boxed{1} \quad \underset{x\rightarrow \infty}{lim}$ $ \dfrac{x+3}{x^{2}{-4}} = 0\\ \\ $(polinom de grad mic pe polinom de grad mai mare, deja putem spune $ \\ $ca limita este 0)[/tex]
[tex]\boxed{2} \quad \underset{x\rightarrow \infty}{lim}$ $ \dfrac{x+3}{x^{2}{-4}} = \underset{x\rightarrow \infty}{lim}$ $ \dfrac{x\cdot\Big(1+ \dfrac{3}{x}\Big) }{x^{2}\cdot\Big(1{-\dfrac{4}{x^2}\Big) }} = \underset{x\rightarrow \infty}{lim}$ $ \dfrac{1\cdot\Big(1+ \dfrac{3}{x}\Big) }{x\cdot\Big(1{-\dfrac{4}{x^2}\Big) }} = [/tex]
[tex]= \dfrac{1\cdot\Big(1+ \dfrac{3}{\infty}\Big) }{\infty\cdot\Big(1{-\dfrac{4}{\infty}\Big) }} = \dfrac{1\cdot\Big(1+ 0\Big) }{\infty\cdot\Big({1-0\Big) }} = \dfrac{1}{\infty} = 0 [/tex]
