Solving a trigonometric limit $\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$ First off, please excuse my n00bishness I have only just begun learning about algebraic manipulation of limits so this is probably a really dumb or obvious question.
I'm trying to solve the following limit:
$$
\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}
$$
This limit is $0/0$ if evaluated directly, so I tried multiplying by the conjugate of the denominator:
$$
\begin{align}
& = \lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{x} + 3)}\\
& = \lim_{x\to\pi/6}\frac{(2\sin{x} - 1)(2\sqrt{3}\cos{x} - 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{3} + 3)} \\
& = \lim_{x\to\pi/6}\frac{2\sin{x} - 1}{2\sqrt{3}\cos{x} + 3}\\
& = \frac{2(1/2) - 1}{2\sqrt{3}\frac{\sqrt{3}}{2} + 3}\\
& = \frac{0}{6}\\
& = 0
\end{align}
$$
But according to WolframAlpha this is incorrect, and the limit should be 1. What have I done wrong?
Also, as I have only just begun I am unfamiliar with L'Hopital's rule.

$$\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$$
$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{\cos x-\cos\frac\pi6}$$
$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{x-\frac\pi6}\cdot\frac1{\lim_{x\to\pi/6}\dfrac{\cos x-\cos\frac\pi6}{x-\frac\pi6}}$$
$$=-\frac2{2\sqrt3}\frac{\dfrac{d(\sin x)}{dx}_{(\text{ at } x=\frac\pi6)}}{\dfrac{d(\cos x)}{dx}_{(\text{ at } x=\frac\pi6)}}$$
$$=\cdots$$
