Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{{x^2} - 4}}\).        b)\[\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} - 5{\rm{x}}} + x} \right)\]

a)          \(\mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{(x - 2)(x + 2)}} = \mathop {\lim }\limits_{x \to 2} \frac{1}{{x + 2}} = \frac{1}{4}\)

b)              \[\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} - 5{\rm{x}}} + x} \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} - 5x - {x^2}}}{{\sqrt {{x^2} - 5x} - x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 5x}}{{\left| x \right|\sqrt {1 - \frac{5}{x}} - x}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 5x}}{{ - x\sqrt {1 - \frac{5}{x}} - x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{5}{{\sqrt {1 - \frac{5}{x}} + 1}} = \frac{5}{2}\end{array}\]