Tính giới hạn I

A

- Hướng dẫn giải

Đáp án A 

Ta có 

I=$lim\left(\sqrt{n^2-2n+3}-n\right)=lim\frac{\left(\sqrt{n^2-2n+3}-n\right)\left(\sqrt{n^2-2n+3}+n\right)}{\sqrt{n^2-2n+3}+n}$

$$\begin{gathered} =lim\frac{\left(n^2-2n+3\right)-n^2}{\sqrt{n^2-2n+3}+n} \\ =lim\frac{-2n+3}{\sqrt{n^2-2n+3}+n} \\ =lim\frac{-2+{\displaystyle \frac{3}{n}}}{\sqrt{1-{\displaystyle \frac{2}{n}}+{\displaystyle \frac{3}{n^2}}}+1} \\ =\frac{-2}{\sqrt{1}+1}=-1 \end{gathered}$$