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D

- Hướng dẫn giải

Đáp án D

$\underset{x\to 0}{\lim }\frac{\sqrt{4x^2-2x+1}-\sqrt{1-2x}}{x}.=\underset{x\to 0}{\lim }\frac{\sqrt{4x^2-2x+1}-1}{x}-\underset{x\to 0}{\lim }\frac{\sqrt{1-2x}}{x}$

$=\underset{x\to 0}{\lim }\frac{4x^2-2x}{\left(\sqrt{4x^2-2x+1}+1\right)}+\underset{x\to 0}{\lim }\frac{2x}{x\left(\sqrt{1-2x}+1\right)}=\underset{x\to 0}{\lim }\frac{4x-2}{\sqrt{4x^2-2x+1}+1}+\underset{x\to 0}{\lim }\frac{2}{\sqrt{1-2x}+1}=-1+1=0$