a. Tìm giới hạn của dãy số cho bởi \({u_n} = n - \sqrt {2{n^2} + n + 1} \).

b. Tim giới hạn: \(\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {4 - x} + \sqrt {x + 1} - x}}{{{x^2} - 9}}\).

a. \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = \mathop {\lim }\limits_{n \to + \infty } \left( {n - \sqrt {2{n^2} + n + 1} } \right) = \mathop {\lim }\limits_{n \to + \infty } n\left( {1 - \sqrt {2 + \frac{1}{n} + \frac{1}{{{n^2}}}} } \right)\)

Vì \(\mathop {\lim }\limits_{n \to + \infty } n = + \infty \) và \(\mathop {\lim }\limits_{n \to + \infty } \left( {1 - \sqrt {2 + \frac{1}{n} + \frac{1}{{{n^2}}}} } \right) = 1 - \sqrt 2 < 0\). Nên \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = - \infty \)

b.

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

\( = \mathop {\mathop {\lim }\limits_{x \to 3} \left( {\frac{{ - 1}}{{\left( {x + 3} \right)\left( {\sqrt {4 - x} + 1} \right)}} + \frac{{ - x}}{{\left( {x + 3} \right)\left( {\sqrt {x + 1} + x - 1} \right)}}} \right)}\limits_{x \to 3} = \frac{{ - 1}}{{12}} + \frac{{ - 1}}{8} = - \frac{5}{{24}}\)