Answer
$\lim\limits_{\Delta x\to0}\dfrac{\sin{(\frac{\pi}{6}+\Delta x)}-\frac{1}{2}}{\Delta x}=\dfrac{\sqrt{3}}{2}.$
Work Step by Step
$\lim\limits_{\Delta x\to0}\dfrac{\sin{(\frac{\pi}{6}+\Delta x)}-\frac{1}{2}}{\Delta x}$
$=\lim\limits_{\Delta x\to0}\dfrac{\sin{\frac{\pi}{6}}\cos{\Delta x}+\cos{\frac{\pi}{6}}\sin{\Delta x}-\frac{1}{2}}{\Delta x}$
$=-\frac{1}{2}(\lim\limits_{\Delta x\to0}\dfrac{1-\cos{\Delta x}}{\Delta x})+\frac{\sqrt{3}}{2}(\lim\limits_{\Delta x\to0}\dfrac{\sin{\Delta x}}{x})$
$=-\frac{1}{2}(0)+\frac{\sqrt{3}}{{2}}(1)=\dfrac{\sqrt{3}}{2}.$
