Đáp án:
\[\lim \frac{{{3^n} - {{4.2}^{n - 1}} - 3}}{{{{3.2}^n} + {4^n}}} = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \frac{{{3^n} - {{4.2}^{n - 1}} - 3}}{{{{3.2}^n} + {4^n}}}\\
= \lim \frac{{{3^n} - {2^{n + 1}} - 3}}{{{{3.2}^n} + {4^n}}}\\
= \lim \frac{{{{\left( {\frac{3}{4}} \right)}^n} - \frac{1}{{{2^{n - 1}}}} - \frac{3}{{{4^n}}}}}{{\frac{3}{{{2^n}}} + {2^n}}}\\
\lim \left( {{{\left( {\frac{3}{4}} \right)}^n} - \frac{1}{{{2^{n - 1}}}} - \frac{3}{{{4^n}}}} \right) = 0\\
\lim \left( {\frac{3}{{{2^n}}} + {2^n}} \right) = + \infty \\
\Rightarrow \lim \frac{{{{\left( {\frac{3}{4}} \right)}^n} - \frac{1}{{{2^{n - 1}}}} - \frac{3}{{{4^n}}}}}{{\frac{3}{{{2^n}}} + {2^n}}} = 0 \Rightarrow \lim \frac{{{3^n} - {{4.2}^{n - 1}} - 3}}{{{{3.2}^n} + {4^n}}} = 0
\end{array}\)
