Giới hạn \(\lim\limits_{x\to+\infty}\dfrac{3-2x}{\sqrt{x^2+5}}\) bằng
\(2\) | |
\(-2\) | |
\(+\infty\) | |
\(-\infty\) |
Chọn phương án B.
\(\begin{aligned}
\lim\limits_{x\to+\infty}\dfrac{3-2x}{\sqrt{x^2+5}}&=\lim\limits_{x\to+\infty}\dfrac{x\left(\dfrac{3}{x}-2\right)}{x\sqrt{1+\dfrac{5}{x^2}}}\\
&=\lim\limits_{x\to+\infty}\dfrac{\dfrac{3}{x}-2}{\sqrt{1+\dfrac{5}{x^2}}}\\
&=\dfrac{0-2}{\sqrt{1+0}}=-2.
\end{aligned}\)