Tính giới hạn \(\mathop {lim}\limits_{x \to  + \in...

A

- Hướng dẫn giải

Ta có \(\frac{1}{{A_k^2}} = \frac{1}{{k\left( {k - 1} \right)}} = \frac{1}{{k - 1}} - \frac{1}{k},\) do đó

\(\frac{1}{{A_n^2}} + \frac{1}{{A_n^2}} + \frac{1}{{A_n^2}} + ... + \frac{1}{{A_n^2}} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{n - 1}} + \frac{1}{n} = 1 - \frac{1}{n}\)

Vậy \(\mathop {lim}\limits_{x \to  + \infty } \left( {\frac{1}{{A_n^2}} + \frac{1}{{A_n^2}} + \frac{1}{{A_n^2}} + ... + \frac{1}{{A_n^2}}} \right) = \mathop {lim}\limits_{x \to  + \infty } \left( {1 - \frac{1}{n}} \right) = 1\)