A. \( + \infty \)
B. \(\dfrac{1}{8}\)
C. -2
D. 1
B
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 4} - 2}}{{2x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {x + 4} - 2} \right)\left( {\sqrt {x + 4} + 2} \right)}}{{2x\left( {\sqrt {x + 4} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{x + 4 - 4}}{{2x\left( {\sqrt {x + 4} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{2x\left( {\sqrt {x + 4} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{2x\left( {\sqrt {x + 4} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{2\left( {\sqrt {x + 4} + 2} \right)}}\\ = \dfrac{1}{{2\left( {\sqrt 4 + 2} \right)}} = \dfrac{1}{8}\end{array}\)
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