Evaluate the limit $LaTeX: \displaystyle \lim_{x \to \infty}\frac{4 x^{2} + 4 x - 4}{2 x^{2} + 6 x + 3}$

The limit is an indeterminate form of the type $LaTeX: \displaystyle \frac{\infty}{\infty}$ . Using L'Hospitial's rule 2 times gives: $LaTeX: \lim_{x \to \infty}\frac{4 x^{2} + 4 x - 4}{2 x^{2} + 6 x + 3} = \lim_{x \to \infty}\frac{8 x + 4}{4 x + 6} = \lim_{x \to \infty}\frac{8}{4} = 2$