Ngân hàng bài tập
C
Tính giới hạn $C=\lim\limits_{x\to+\infty}\left(\sqrt{x^2-x+1}-x\right)$.
 | $C=+\infty$ |
 | $C=-\infty$ |
 | $C=\dfrac{1}{2}$ |
 | $C=-\dfrac{1}{2}$ |
1 lời giải
Chọn phương án D.
$\begin{aligned}
C&=\lim\limits_{x\to+\infty}\dfrac{\left(\sqrt{x^2-x+1}-x\right)\left(\sqrt{x^2-x+1}+x\right)}{\sqrt{x^2-x+1}+x}\\
&=\lim\limits_{x\to+\infty}\dfrac{\big(x^2-x+1\big)-x^2}{\sqrt{x^2\left(1-\dfrac{1}{x}+\dfrac{1}{x^2}\right)}+x}\\
&=\lim\limits_{x\to+\infty}\dfrac{-x+1}{x\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}+x}\\
&=\lim\limits_{x\to+\infty}\dfrac{-1+\dfrac{1}{x}}{\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}+1}\\
&=\dfrac{-1+0}{\sqrt{1-0+0}+1}=-\dfrac{1}{2}.
\end{aligned}$