Tính: a) Lim căn {{n^2} + 5n}  - n

a) Ta có \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 5n} - n} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} + 5n - {n^2}}}{{\sqrt {{n^2} + 5n} + n}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{5n}}{{\sqrt {{n^2} + 5n} + n}}\)

                                                =\(\mathop {\lim }\limits_{n \to + \infty } \frac{5}{{\sqrt {1 + \frac{5}{n}} + 1}} = \frac{5}{2}\).

b) Ta có \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 6} \right) = 8 > 0\). \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 2} \right) = 0\) và \(x - 2 < 0,\forall x \to {2^ - }\)

Vậy \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 6}}{{x - 2}} = - \infty \)