Tính \(L=\lim\left(\sqrt{n^2-n+1}-n\right)\).
\(-\dfrac{1}{2}\) | |
\(0\) | |
\(1\) | |
\(-\infty\) |
Chọn phương án A.
Dùng máy tính cầm tay:
Chọn phương án A.
\(\begin{aligned}
L&=\lim\left(\sqrt{n^2-n+1}-n\right)\\
&=\lim\dfrac{\left(\sqrt{n^2-n+1}-n\right)\left(\sqrt{n^2-n+1}+n\right)}{\sqrt{n^2-n+1}+n}\\
&=\lim\dfrac{\left(n^2-n+1\right)-n^2}{\sqrt{n^2\left(1-\dfrac{1}{n}+\dfrac{1}{n^2}\right)}+n}\\
&=\lim\dfrac{-n+1}{n\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+n}\\
&=\lim\dfrac{-1+\dfrac{1}{n}}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\\
&=\dfrac{-1+0}{\sqrt{1-0+0}+1}=-\dfrac{1}{2}.
\end{aligned}\)