Tính các giới hạn sau

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

\[ = \mathop {\lim }\limits_{n \to + \infty } \frac{{1 - \frac{1}{n}}}{{\sqrt {1 + \frac{1}{n}} + \sqrt {1 + \frac{1}{{{n^2}}}} }} = \frac{1}{{1 + 1}} = \frac{1}{2}.\]

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