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Câu hỏi: $\underset{x\to +\infty }{\mathop{\lim }} \dfrac{x+3}{\sqrt{4{{x}^{2}}+1}-2}$ bằng
A. $\dfrac{1}{4}$
B. $\dfrac{1}{2}$
C. $-\dfrac{3}{2}$
D. 0
A. $\dfrac{1}{4}$
B. $\dfrac{1}{2}$
C. $-\dfrac{3}{2}$
D. 0
Ta có $L=\underset{x\to +\infty }{\mathop{\lim }} \dfrac{x\left( 1+\dfrac{3}{x} \right)}{x\sqrt{4+\dfrac{1}{{{x}^{2}}}}-2}=\underset{x\to +\infty }{\mathop{\lim }} \dfrac{1+\dfrac{3}{x}}{x\sqrt{4+\dfrac{1}{{{x}^{2}}}}-\dfrac{2}{x}}=\dfrac{1}{2}$
Đáp án B.