Question

CMR \(lim\dfrac{n^2}{2^n}\)=0

Answer 1

\(2^n=\left(1+1\right)^2=1+C_n^1+C_n^2+C_n^3+...+C_n^n>C_n^3\) (khi n đủ lớn)

\(\Rightarrow2^n>\dfrac{n\left(n-1\right)\left(n-2\right)}{6}\)

\(\Rightarrow\dfrac{n^2}{2^n}< \dfrac{6n^2}{n\left(n-1\right)\left(n-2\right)}=\dfrac{6n}{\left(n-1\right)\left(n-2\right)}\)

Đồng thời do \(\left\{{}\begin{matrix}n^2>0\\2^n>0\end{matrix}\right.\) \(\Rightarrow\dfrac{n^2}{2^n}>0\)

\(\Rightarrow0< \dfrac{n^2}{2^n}< \dfrac{6n}{\left(n-1\right)\left(n-2\right)}\)

Mà \(\lim\left(0\right)=\lim\left(\dfrac{6n}{\left(n-1\right)\left(n-2\right)}\right)=0\)

\(\Rightarrow\lim\left(\dfrac{n^2}{2^n}\right)=0\)