Answer
$\dfrac {1}{8}=0.125 $
Work Step by Step
$\lim _{h\rightarrow 0}\dfrac {\sqrt {16+h}-4}{h}=\dfrac {\sqrt {16+h}-4}{h}\times \dfrac {\sqrt {16+h}+4}{\sqrt {16+h}+4}=\dfrac {\left( \sqrt {16+h}\right) ^{2}-4^{2}}{h\times \left( \sqrt {16+h}+4\right) }=\dfrac {16+h-16}{h\times \left( \sqrt {16+h}+4\right) }=\dfrac {1}{\sqrt {16+h}+4}=\dfrac {1}{\sqrt {16}+4}=\dfrac {1}{8} $
