Tính \(L=\lim\left(\sqrt{n^2+2n}-\sqrt{n^2-2n}\right)\).
\(1\) | |
\(2\) | |
\(4\) | |
\(+\infty\) |
Chọn phương án B.
Dùng máy tính cầm tay:
Chọn phương án B.
\(\begin{aligned}
L&=\lim\left(\sqrt{n^2+2n}-\sqrt{n^2-2n}\right)\\
&=\lim\dfrac{\left(\sqrt{n^2+2n}-\sqrt{n^2-2n}\right)\left(\sqrt{n^2+2n}+\sqrt{n^2-2n}\right)}{\sqrt{n^2+2n}+\sqrt{n^2-2n}}\\
&=\lim\dfrac{\left(n^2+2n\right)-\left(n^2-2n\right)}{\sqrt{n^2\left(1+\dfrac{2}{n}\right)}+\sqrt{n^2\left(1-\dfrac{2}{n}\right)}}\\
&=\lim\dfrac{4n}{n\sqrt{1+\dfrac{2}{n}}+n\sqrt{1-\dfrac{2}{n}}}\\
&=\lim\dfrac{4}{\sqrt{1+\dfrac{2}{n}}+\sqrt{1-\dfrac{2}{n}}}\\
&=\dfrac{4}{\sqrt{1+0}+\sqrt{1-0}}=2.
\end{aligned}\)