Tính giới hạn sau: \(\mathop {\lim }\limits_{x \to...

B

- Hướng dẫn giải

$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 4}  - 2}}{{2x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {x + 4}  - 2} \right)\left( {\sqrt {x + 4}  + 2} \right)}}{{2x\left( {\sqrt {x + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{x + 4 - 4}}{{2x\left( {\sqrt {x + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{2x\left( {\sqrt {x + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{2x\left( {\sqrt {x + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{2\left( {\sqrt {x + 4}  + 2} \right)}}\\ = \dfrac{1}{{2\left( {\sqrt 4  + 2} \right)}} = \dfrac{1}{8}\end{array}$