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Câu hỏi: Tính $\underset{x\to +\infty }{\mathop{\lim }} \dfrac{x+3}{\sqrt{4{{x}^{2}}+1}-2}$
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 $L=\underset{x\to +\infty }{\mathop{\lim }} \dfrac{1+\dfrac{3}{x}}{\sqrt{4+\dfrac{1}{{{x}^{2}}}}-\dfrac{2}{x}}=\dfrac{1+0}{\sqrt{4}-0}=\dfrac{1}{2}.$ Chọn C.
Đáp án C.