Tính giới hạn \(\lim\limits_{x\to3^-}\dfrac{3-x}{\sqrt{27-x^3}}\).
\(\dfrac{1}{3}\) | |
\(0\) | |
\(\dfrac{5}{3}\) | |
\(\dfrac{3}{5}\) |
Chọn phương án B.
Với \(x\to3^-\) thì \(x<3\). Suy ra \(3-x>0\).
Khi đó $$\begin{aligned}
\lim\limits_{x\to3^-}\dfrac{3-x}{\sqrt{27-x^3}}&=\lim\limits_{x\to3^-}\dfrac{3-x}{\sqrt{(3-x)\left(9+3x+x^2\right)}}\\
&=\lim\limits_{x\to3^-}\dfrac{\sqrt{3-x}}{\sqrt{9+3x+x^2}}\\
&=\dfrac{\sqrt{3-3}}{\sqrt{9+3\cdot3+3^2}}=0.
\end{aligned}$$