Tìm các giới hạn sau

a. \(\mathop {\lim }\limits_{x \to 2} \frac{{x + 2}}{x}\). b. \(\mathop {\lim }\limits_{x \to - \infty } \frac{x}{{x + 1}}\) c. \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {5x - 1} - x - 1}}{{{x^2} - 3x + 2}}\)

a) \(\mathop {\lim }\limits_{x \to 2} \frac{{x + 2}}{x} = 2\).

b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{x}{{x + 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{x}{{x\left( {1 + \frac{1}{x}} \right)}} = 1\).

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {5x - 1} - x - 1}}{{{x^2} - 3x + 2}} = \mathop {\lim }\limits_{x \to 1} \frac{{5x - 1 - {{\left( {x + 1} \right)}^2}}}{{\left( {{x^2} - 3x + 2} \right)\left( {\sqrt {5x - 1} + x + 1} \right)}}\)

\[ = \mathop {\lim }\limits_{x \to 1} \frac{{ - {x^2} + 3x - 2}}{{\left( {{x^2} - 3x + 2} \right)\left( {\sqrt {5x - 1} + x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - 1}}{{\left( {\sqrt {5x - 1} + x + 1} \right)}} = - \frac{1}{4}\]