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