Evaluate the limit LaTeX: \displaystyle \lim_{x \to \infty}\frac{8 x^{2} - 6 x + 2}{8 x^{2} + 2 x - 5}

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{8 x^{2} - 6 x + 2}{8 x^{2} + 2 x - 5} = \lim_{x \to \infty}\frac{16 x - 6}{16 x + 2} = \lim_{x \to \infty}\frac{16}{16} = 1