Answer
$ \dfrac{1}{8}$
Work Step by Step
In order to to find the solution we will rationalize the function and then simplify.
$\dfrac{\sqrt {16+x}-4}{x} \times \dfrac{\sqrt {16+x}+4}{\sqrt {16+x}+4}$
or, $=\dfrac{(16+x)-16}{x[\sqrt {16+x}+4]}$
or, $=\dfrac{1}{\sqrt {16+x}+4}$
Now, $\lim_\limits{x\to 0} \dfrac{\sqrt {16+x}-4}{x}=\lim_\limits{x\to 0} \dfrac{1}{\sqrt {16+x}+16}$
or, $= \dfrac{1}{\sqrt {16}+4}$
or, $= \dfrac{1}{8}$
