Tính giới hạn sau: lim 2 n + 2 + 4 ⋅ 6 n − 1 + 2 3 n + 1 + 6 n − 1 + 1 .

\(\lim \frac{{{2^{n + 2}} + 4 \cdot {6^{n - 1}} + 2}}{{{3^{n + 1}} + {6^{n - 1}} + 1}} = \lim \frac{{4{{\left( {\frac{1}{3}} \right)}^n} + \frac{2}{3} + 2 \cdot {{\left( {\frac{1}{6}} \right)}^n}}}{{3 \cdot {{\left( {\frac{1}{2}} \right)}^n} + \frac{1}{6} + {{\left( {\frac{1}{6}} \right)}^n}}} = \frac{{4 \cdot 0 + \frac{2}{3} + 2 \cdot 0}}{{3 \cdot 0 + \frac{1}{6} + 0}} = 4\).

Trả lời: 4.