How to find $\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$ without using L'Hospital's Rule? I have to find $$\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$$
We are not allowed to use L'Hospital's rule, any suggestions would be beneficial!
I have tried multiplying by the conjugate (both the numerator and the denominator).
I have tried using trig substitution. 

Using the definition of the derivative of $\sin(x)$ at $x = \dfrac{\pi}{6}$, we get
\begin{align}
   \lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}
   &=\lim_{x \to \frac{\pi}{6}}
     \frac{\sin(x)-\sin(\frac{\pi}{6})}{x-\frac{\pi}{6}} \\
   &= \sin'(\frac{\pi}{6}) \\
   &= \cos(\frac{\pi}{6}) \\
   &= \dfrac{\sqrt 3}{2}
\end{align}
