Tính các giới hạn sau

                a) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{2{n^2} - n - 2}}{{{n^2} + n}}\)       b) \[\mathop {\lim }\limits_{x \to 3} \frac{{2 - \sqrt {x + 1} }}{{{x^2} - 9}}.\]         c) \[\mathop {\lim }\limits_{x \to 1} \left( {\frac{{13}}{{1 - {x^{13}}}} - \frac{1}{{1 - x}}} \right).\]

    a) Ta có \(\mathop {\lim }\limits_{n \to + \infty } \frac{{2{n^2} - n - 2}}{{{n^2} + n}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{2 - \frac{1}{n} - 2.\frac{1}{{{n^2}}}}}{{1 + \frac{1}{n}}} = \frac{{2 - 0 - 0}}{{1 + 0}} = 2\).

           b) Ta có

 \[\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \frac{{2 - \sqrt {x + 1} }}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \frac{{4 - (x + 1)}}{{(x - 3)(x + 3)(2 + \sqrt {x + 1} )}} = \mathop {\lim }\limits_{x \to 3} \frac{{ - (x - 3)}}{{(x - 3)(x + 3)(2 + \sqrt {x + 1} )}}\\ = \mathop {\lim }\limits_{x \to 3} \frac{{ - 1}}{{(x + 3)(2 + \sqrt {x + 1} )}} = \frac{{ - 1}}{{24}}\\\end{array}\]

                c) Ta có

                \[\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left( {\frac{{13}}{{1 - {x^{13}}}} - \frac{1}{{1 - x}}} \right) = \mathop {\lim }\limits_{x \to 1} \frac{{13 - (1 + x + {x^2} + ... + {x^{12}})}}{{1 - {x^{13}}}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{12 - x - {x^2} - ... - {x^{12}}}}{{1 - {x^{13}}}} = \mathop {\lim }\limits_{x \to 1} \frac{{1 - x + 1 - {x^2} + ... + 1 - {x^{12}}}}{{1 - {x^{13}}}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{(1 - x)\left[ {1 + (1 + x) + (1 + x + {x^2}) + ... + (1 + x + {x^2} + ... + {x^{11}})} \right]}}{{(1 - x)(1 + x + {x^2} + ... + {x^{12}})}}\end{array}\]

                \[\begin{array}{l} = \mathop {\lim }\limits_{x \to 1} \frac{{1 + (1 + x) + (1 + x + {x^2}) + ... + (1 + x + {x^2} + ... + {x^{11}})}}{{(1 + x + {x^2} + ... + {x^{12}})}}\\ = \frac{{1 + 2 + 3 + ... + 12}}{{13}} = \frac{{13.6}}{{13}} = 6\end{array}\]