Tính các giới hạn sau:

a) \[\mathop {\lim }\limits_{n \to + \infty } \left( {1 + n - {n^2}} \right).\]  b) \[\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 8}}{{{x^2} - 4}}.\]

a) Ta có: \[1 + n - {n^2} = {n^2}\left( {\frac{1}{{{n^2}}} + \frac{1}{n} - 1} \right).\]

Mặt khác: \(\mathop {\lim }\limits_{n \to + \infty } {n^2} = + \infty ;\)

                 \(\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{{n^2}}} + \frac{1}{n} - 1} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{{n^2}}} + \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n} - \mathop {\lim }\limits_{n \to + \infty } 1 = 0 + 0 - 1 = - 1 < 0.\)

\( \Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {1 + n - {n^2}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left[ {{n^2}\left( {\frac{1}{{{n^2}}} + \frac{1}{n} - 1} \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } {n^2}.\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{{n^2}}} + \frac{1}{n} - 1} \right) = - \infty .\)

b) \[\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 8}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 2x + 4}}{{x + 2}} = \frac{{{2^2} + 2.2 + 4}}{{2 + 2}} = 3.\]