Tính các giới hạn sau: (a) lim n → + ∞ √ n ( √ n + 1 − √ n ) ;

a) \[\mathop {\lim }\limits_{n \to + \infty } \sqrt n \left( {\sqrt {n + 1} - \sqrt n } \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt n \left( {\sqrt {n + 1} - \sqrt n } \right)\left( {\sqrt {n + 1} + \sqrt n } \right)}}{{\sqrt {n + 1} + \sqrt n }}\]

\[ = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt n \left( {n + 1 - n} \right)}}{{\sqrt {n + 1} + \sqrt n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt n }}{{\sqrt {n + 1} + \sqrt n }}\]

\[ = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt n }}{{\sqrt n \left( {\sqrt {1 + \frac{1}{n}} + 1} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{\sqrt {1 + \frac{1}{n}} + 1}} = \frac{1}{2}\].

b) Ta có: \[\mathop {\lim }\limits_{x \to 2} \left( {5 - x} \right) = 5 - 2 = 3 > 0;\] \[\mathop {\lim }\limits_{x \to 2} \frac{1}{{{{\left( {x - 2} \right)}^2}}} = + \infty \]

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

a) \[\mathop {\lim }\limits_{n \to + \infty } \sqrt n \left( {\sqrt {n + 1} - \sqrt n } \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt n \left( {\sqrt {n + 1} - \sqrt n } \right)\left( {\sqrt {n + 1} + \sqrt n } \right)}}{{\sqrt {n + 1} + \sqrt n }}\]