Phương pháp giải:

Nếu \({x_n} < {u_n} < {v_n}\)  mà  \(\lim \,\,{x_n} = \lim \,\,{v_n} = a \Rightarrow \lim \,\,{u_n} = a\)

Lời giải chi tiết:

Ta có: \(\frac{k}{{(k + 1)!}} = \frac{1}{{k!}} - \frac{1}{{(k + 1)!}}\) nên \({x_k} = 1 - \frac{1}{{(k + 1)!}}\)

\(\begin{array}{l} \Rightarrow {x_k} - {x_{k + 1}} = 1 - \frac{1}{{\left( {k + 1} \right)!}} - 1 + \frac{1}{{(k + 2)!}} = \frac{1}{{(k + 2)!}} - \frac{1}{{(k + 1)!}} < 0\\ \Rightarrow {x_k} < {x_{k + 1}} \Rightarrow {x_1} < {x_2} < ... < {x_{2011}}\\ \Rightarrow x_1^n < x_2^n < ... < x_{2011}^n \Rightarrow \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}} < \sqrt[n]{{x_{2011}^n + x_{2011}^n + ... + x_{2011}^n}} = \sqrt[n]{{2011.x_{2011}^n}} = \sqrt[n]{{2011}}.{x_{2011}}\end{array}\)

Lại có: \({x_{2011}} = \sqrt[n]{{x_{2011}^n}} < \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}}\)

Vậy: \({x_{2011}} < \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}} < \sqrt[n]{{2011}}{x_{2011}}\)

Ta có: \({x_{2011}} = 1 - \frac{1}{{\left( {2011 + 1} \right)!}} = 1 - \frac{1}{{2012!}}\)

\(\begin{array}{l} \Rightarrow \lim {x_{2011}} = {x_{2011}} = 1 - \frac{1}{{2012!}}\\ \Rightarrow \lim \sqrt[n]{{2011}}{x_{2011}} = \lim \,\,{2011^{\frac{1}{n}}}{x_{2011}} = {2011^0}.{x_{2011}} = {x_{2011}} = 1 - \frac{1}{{2012!}}\end{array}\)

Vậy \(\lim {u_n} = 1 - \frac{1}{{2012!}}\).

Chọn C.