Giải thích các bước giải:

$\lim\limits_{x \to 0}$$\dfrac{\sqrt[]{4+x}-2}{4x}$

$=\lim\limits_{x \to 0}$$\dfrac{4+x-4}{4x(\sqrt[]{4+x}+2)}$

$=\lim\limits_{x \to 0}$$\dfrac{1}{4(\sqrt[]{4+x}+2)}$

$=\dfrac{\lim\limits_{x \to 0}1}{\lim\limits_{x \to 0}(4.(\sqrt[]{4+x}+2))}$

$=\dfrac{1}{4(\sqrt[]{4+0}+2)}$$=\dfrac{1}{16}$