A. \(+\infty\)
B. \(-\infty\)
C. \(\frac{2}{3}\)
D. \(\frac{3}{2}\)
A
Ta có \(2^{n}=\sum_{k=0}^{n} C_{n}^{k} \Rightarrow 2^{n} \geq C_{n}^{3}=\frac{n(n-1)(n-2)}{6} \sim \frac{n^{3}}{6} \Rightarrow\left\{\begin{array}{l} \frac{n}{2^{n}} \rightarrow 0 \\ \frac{2^{n}}{n^{2}} \rightarrow+\infty \end{array}\right.\)
Khi đó
\(\lim \frac{2^{n+1}+3 n+10}{3 n^{2}-n+2}=\lim \frac{2^{n}}{n^{2}} \cdot \frac{2+3 \cdot \frac{n}{2^{n}}+10 \cdot\left(\frac{1}{2}\right)^{n}}{3-\frac{1}{n}+\frac{2}{n^{2}}}=+\infty\)
vì
\(\left\{\begin{array}{l} \lim \frac{2^{n}}{n^{2}}=+\infty \\ \lim \frac{2+3 \cdot \frac{n}{2^{n}}+10 \cdot\left(\frac{1}{2}\right)^{n}}{3-\frac{1}{n}+\frac{2}{n^{2}}}=\frac{2}{3}>0 \end{array}\right.\)
Câu hỏi trên thuộc đề trắc nghiệm dưới đây !
Copyright © 2021 HOCTAP247