Cho dãy số

- Hướng dẫn giải

Xét tử số : $\sqrt{4n^2+1}-2n=\frac{4n^2+1-4n^2}{\sqrt{4n^2+1}+2n}=\frac{1}{\sqrt{4n^2+1}+2n}$

Xét mẫu số: $\frac{1}{\sqrt{n^2+4n+1}-n}=\frac{\sqrt{n^2+4n+1}+\text{}n}{n^2+4n+1-n^2}=\frac{\sqrt{n^2+4n+1}+n}{4n+1}$

$$\begin{gathered} \Rightarrow lim{u}_n=\lim \frac{\sqrt{n^2+4n+1}+n}{\left(\sqrt{4n^2+1}+2n\right)\left(2n+1\right)} \\ =\lim \frac{\sqrt{n^2\left(1+\frac{4}{n}+\frac{1}{n^2}\right)}+n}{\left(\sqrt{n^2\left(4+\frac{1}{n^2}\right)}+2n\right)\left(2n+1\right)} \\ =\lim \frac{n\left[\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+1\right]}{n\left(\sqrt{4+\frac{1}{n^2}}+2\right)n\left(2+\frac{1}{n}\right)}=\lim \frac{\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+1}{n\left(\sqrt{4+\frac{1}{n^2}}+2\right)\left(2+\frac{1}{n}\right)} \\ =\lim \frac{2}{n\left(2+2\right)2}=\lim \frac{1}{4n}=0. \end{gathered}$$ 

Do đó $\lim u_n=0$

Chọn đáp án C