Câu hỏi 37

\(\begin{array}{l}f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sqrt {{x^2} + 1}  - 1}}{x}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1}  - 1}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + 1 - 1}}{{{x^2}\left( {\sqrt {{x^2} + 1}  + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {{x^2} + 1}  + 1}} = \frac{1}{2}\end{array}\)

Chọn A.