Ngân hàng bài tập
SS
Cho $\lim\limits_{x\to-\infty}\left(\sqrt{ax^2-2x}+bx\right)=11$. Tính $Q=b-a$.
 | $Q=\dfrac{17}{121}$ |
 | $Q=\dfrac{5}{121}$ |
 | $Q=-\dfrac{13}{121}$ |
 | $Q=\dfrac{10}{121}$ |
1 lời giải
Chọn phương án D.
$\begin{aligned}
\lim\limits_{x\to-\infty}\left(\sqrt{ax^2-2x}+bx\right)&=\lim\limits_{x\to-\infty}\dfrac{\left(\sqrt{ax^2-2x}+bx\right)\left(\sqrt{ax^2-2x}-bx\right)}{\sqrt{ax^2-2x}-bx}\\
&=\lim\limits_{x\to-\infty}\dfrac{\big(ax^2-2x\big)-b^2x^2}{\sqrt{x^2\left(a-\dfrac{2}{x}\right)}-bx}\\
&=\lim\limits_{x\to-\infty}\dfrac{\big(a-b^2\big)x^2-2x}{|x|\sqrt{a-\dfrac{2}{x}}-bx}\\
&=\lim\limits_{x\to-\infty}\dfrac{\big(a-b^2\big)x^2-2x}{-x\sqrt{a-\dfrac{2}{x}}-bx}\\
&=\lim\limits_{x\to-\infty}\dfrac{\big(b^2-a\big)x+2}{\sqrt{a-\dfrac{2}{x}}+b}.
\end{aligned}$
Vì $\lim\limits_{x\to-\infty}\dfrac{\big(b^2-a\big)x+2}{\sqrt{a-\dfrac{2}{x}}+b}$ hữu hạn nên $b^2-a=0\Leftrightarrow b=\sqrt{a}$.
Ngoài ra $\begin{aligned}[t]
\lim\limits_{x\to-\infty}\dfrac{\big(b^2-a\big)x+2}{\sqrt{a-\dfrac{2}{x}}+b}=11&\Leftrightarrow\dfrac{2}{\sqrt{a-0}+b}=11\\
&\Leftrightarrow\sqrt{a}+b=\dfrac{2}{11}\\
&\Leftrightarrow2b=\dfrac{2}{11}\\
&\Leftrightarrow b=\dfrac{1}{11}\\
&\Rightarrow a=b^2=\dfrac{1}{121}.
\end{aligned}$
Vậy $Q=b-a=\dfrac{10}{121}$.