a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{1 - 3{x^2}}}{{{x^2} + 2x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}\left( {\frac{1}{{{x^2}}} - 3} \right)}}{{{x^2}\left( {1 + \frac{{2x}}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{{{x^2}}} - 3}}{{1 + \frac{2}{x}}} = \frac{{\mathop {\lim }\limits_{x \to + \infty } \frac{1}{{{x^2}}} - \mathop {\lim }\limits_{x \to + \infty } 3}}{{\mathop {\lim }\limits_{x \to + \infty } 1 + \mathop {\lim }\limits_{x \to + \infty } \frac{2}{x}}} = \frac{{0 - 3}}{{1 + 0}} = - 3\)
b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{2}{{x + 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{2}{{x\left( {1 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}.\mathop {\lim }\limits_{x \to - \infty } \frac{2}{{1 + \frac{1}{x}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}.\frac{{\mathop {\lim }\limits_{x \to - \infty } 2}}{{\mathop {\lim }\limits_{x \to - \infty } 1 + \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}}} = 0.\frac{2}{{1 + 0}} = 0\).