Question

Answer 1

Chọn B

Answer 2

\(S=\dfrac{1}{\sqrt{4n^2+1}}+\dfrac{1}{\sqrt{4n^2+2}}+...+\dfrac{1}{\sqrt{4n^2+n}}>\dfrac{1}{\sqrt{4n^2+n}}+...+\dfrac{1}{\sqrt{4n^2+n}}=\dfrac{n}{\sqrt{4n^2+n}}\)

\(\dfrac{1}{\sqrt{4n^2+1}}+...+\dfrac{1}{\sqrt{4n^2+n}}< \dfrac{1}{\sqrt{4n^2+1}}+...+\dfrac{1}{\sqrt{4n^2+1}}=\dfrac{n}{\sqrt{4n^2+1}}\)

Mà \(\lim\dfrac{n}{\sqrt{4n^2+n}}=\lim\dfrac{n}{\sqrt{4n^2+1}}=\dfrac{1}{2}\)

\(\Rightarrow\lim\left(S\right)=\dfrac{1}{2}\) theo định lý kẹp