Answer
$\lim\limits_{\Delta x\to0}\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}=\dfrac{1}{2\sqrt{x}}.$
Work Step by Step
$\lim\limits_{\Delta x\to0}\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$
$=\lim\limits_{\Delta x\to0}\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\times\dfrac{\sqrt{x+\Delta x}+\sqrt{x}}{\sqrt{x+\Delta x}+\sqrt{x}}$
$=\lim\limits_{\Delta x\to0}\dfrac{(\sqrt{x+\Delta x})^2-(\sqrt{x})^2}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}$
$=\lim\limits_{\Delta x\to0}\dfrac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}$
$=\lim\limits_{\Delta x\to0}\dfrac{1}{\sqrt{x+\Delta x}+\sqrt{x}}$
$=\dfrac{1}{\sqrt{x+0}+\sqrt{x}}=\dfrac{1}{2\sqrt{x}}.$
