Đề bài

Cho \({u_n} = \sqrt n \left( {\sqrt {n + 2}  - \sqrt {n - 1} } \right)\). Khi đó \(\mathop {\lim }\limits_{n \to  + \infty } {u_n}\) bằng

A.\( + \infty \)                         

B. 0                     

C. \(\frac{1}{2}\)                    

D. 1.

Lời giải chi tiết

\(\begin{array}{l}\mathop {\lim }\limits_{n \to  + \infty } {u_n} = \mathop {\lim }\limits_{n \to  + \infty } \sqrt n \left( {\sqrt {n + 2}  - \sqrt {n + 1} } \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt n \left( {\sqrt {n + 2}  - \sqrt {n + 1} } \right)\left( {\sqrt {n + 2}  + \sqrt {n + 1} } \right)}}{{\left( {\sqrt {n + 2}  + \sqrt {n + 1} } \right)}}\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt n \left( {n + 2 - n - 1} \right)}}{{\sqrt {n + 2}  + \sqrt {n - 1} }} = \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt n }}{{\sqrt {n + 2}  + \sqrt {n + 1} }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{\sqrt {1 + \frac{2}{n}}  + \sqrt {1 + \frac{2}{n}} }} = \frac{1}{2}\end{array}\)Đáp án C