Tính các giới hạn sau

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

b) \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} - x} + 2x} \right)\)

\(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} - x} + 2x} \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{4{x^2} - x - 4{x^2}}}{{\sqrt {4{x^2} - x} - 2x}}\)

\( = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x}}{{ - x\sqrt {4 - \frac{1}{x}} - 2x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 1}}{{ - \sqrt {4 - \frac{1}{x}} - 2}} = \frac{1}{4}\).