Đáp án:
\[\lim {u_n} = 4\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{u_{n + 1}} = \frac{{5{u_n} + 4}}{{{u_n} + 2}}\\
\Leftrightarrow {u_{n + 1}} = \frac{{4\left( {{u_n} + 2} \right) + \left( {{u_n} - 4} \right)}}{{{u_n} + 2}}\\
\Leftrightarrow {u_{n + 1}} = 4 + \frac{{{u_n} - 4}}{{{u_n} + 2}}\\
\Leftrightarrow {u_{n + 1}} - 4 = \frac{{{u_n} - 4}}{{{u_n} + 2}}\\
\Leftrightarrow \frac{1}{{{u_{n + 1}} - 4}} = \frac{{{u_n} + 2}}{{{u_n} - 4}}\\
\Leftrightarrow \frac{1}{{{u_{n + 1}} - 4}} = \frac{{\left( {{u_n} - 4} \right) + 6}}{{{u_n} - 4}}\\
\Leftrightarrow \frac{1}{{{u_{n + 1}} - 4}} = 1 + \frac{6}{{{u_n} - 4}}\\
{x_n} = \frac{1}{{{u_n} - 4}} \Rightarrow {x_{n + 1}} = 1 + 6{x_n} \Leftrightarrow {x_{n + 1}} + \frac{1}{5} = \frac{6}{5} + 6.{x_n} = 6.\left( {{x_n} + \frac{1}{5}} \right)\\
{v_n} = {x_n} + \frac{1}{5} \Rightarrow \left\{ \begin{array}{l}
{v_1} = \frac{1}{{{u_1} - 4}} + \frac{1}{5} = \frac{6}{5}\\
{v_{n + 1}} = 6.{v_n}
\end{array} \right.
\end{array}\)
Do đó, \(\left( {{v_n}} \right)\) là CSN có \({v_1} = \frac{6}{5};\,\,\,q = 6\)
Suy ra:
\(\begin{array}{l}
{v_n} = {v_1}.{q^{n - 1}} = \frac{6}{5}{.6^{n - 1}} = \frac{{{6^n}}}{5} \Rightarrow {x_n} = {v_n} - \frac{1}{5} = \frac{{{6^n} - 1}}{5}\\
\Rightarrow {u_n} - 4 = \frac{1}{{{x_n}}} = \frac{5}{{{6^n} - 1}} \Rightarrow {u_n} = 4 + \frac{5}{{{6^n} - 1}}\\
\lim {u_n} = \lim \left( {4 + \frac{5}{{{6^n} - 1}}} \right)\\
\lim \left( {{6^n} - 1} \right) = + \infty \Rightarrow \lim \frac{5}{{{6^n} - 1}} = 0 \Rightarrow \lim {u_n} = \lim \left( {4 + \frac{5}{{{6^n} - 1}}} \right) = 4
\end{array}\)
