Tính giới hạn cho sau: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 4} - 2}}{{2x}}\)

Câu hỏi :

Tính giới hạn sau: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 4}  - 2}}{{2x}}\)

A. \( + \infty \)  

B. \(\dfrac{1}{8}\) 

C. -2   

D.

* Đáp án

B

* Hướng dẫn giải

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

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