Answer
$\dfrac{\sqrt7}{14}$
Work Step by Step
$\lim\limits_{h\to0}\dfrac{f(x+h)f(x)}{h}=\lim\limits_{h\to0}\dfrac{h}{h(\sqrt{x+h}+\sqrt x)}=\lim\limits_{h\to0}\dfrac{1}{(\sqrt{x+h}+\sqrt x)}$
Plug $x=7$ we get
$\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\dfrac{1}{\sqrt{7+h}+\sqrt7}=\dfrac{1}{\sqrt7+\sqrt7}$
and $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2\sqrt7}$
Multiply both numerator and denominator with $\sqrt7$, we have
$\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\dfrac{\sqrt7}{2\times7}=\dfrac{\sqrt7}{14}$
