Tính các giới hạn sau

a) \[\lim \frac{{8n + 5}}{{2n - 1}}\]\[ = \lim \frac{{8 + \frac{5}{n}}}{{2 - \frac{1}{n}}} = \frac{{8 + 0}}{{2 - 0}} = 4\]

              b) \[\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^2} + 1}}{{1 - {x^2}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2}(2 + \frac{1}{{{x^2}}})}}{{{x^2}(\frac{1}{{{x^2}}} - 1)}}\]\[ = \mathop {\lim }\limits_{x \to - \infty } \frac{{2 + \frac{1}{{{x^2}}}}}{{\frac{1}{{{x^2}}} - 1}} = \frac{{2 + 0}}{{0 - 1}} = - 2\]

              c) \[\mathop {\lim }\limits_{x \to \sqrt 2 } \frac{{3{x^2} - 6}}{{x - \sqrt 2 }} = \mathop {\lim }\limits_{x \to \sqrt 2 } \frac{{3(x - \sqrt 2 )(x + \sqrt 2 )}}{{x - \sqrt 2 }}\]=\[\mathop {\lim }\limits_{x \to \sqrt 2 } 3(x + \sqrt 2 ) = 6\sqrt 2 \].