Answer
$\frac{1}{32}$
Work Step by Step
The limit can be written:
$$\begin{align*}
\lim\limits_{h \to 0}\frac{\sqrt[4] {16+h}-2}{h}& =\lim\limits_{h \to 0}\frac{\sqrt[4] {16+h}-\sqrt[4]{16}}{h} \\
&=\left[\frac{d}{dx}\sqrt[4] x\right]_{x=16}\\
& = \frac{1}{4}x^{-\frac{3}{4}}|_{x=16}\\
& = \frac{1}{32}
\end{align*}$$