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Câu hỏi: $\underset{x\to 1}{\mathop{\lim }} \dfrac{\sqrt{x+3}-2}{x-1}$ bằng
A. $+\infty $.
B. 1.
C. $\dfrac{1}{2}$.
D. $\dfrac{1}{4}$.
A. $+\infty $.
B. 1.
C. $\dfrac{1}{2}$.
D. $\dfrac{1}{4}$.
$\underset{x\to 1}{\mathop{\lim }} \dfrac{\sqrt{x+3}-2}{x-1}=\underset{x\to 1}{\mathop{\lim }} \dfrac{\dfrac{x+3-1}{\sqrt{x+3}+2}}{x-1}=\underset{x\to 1}{\mathop{\lim }} \dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}$.
Đáp án D.