<s>[INST] 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.
 [/INST] As $\displaystyle\sin x-\sin a=\frac{\sin^2x-\sin^2a}{\sin x+\sin a}$ 
using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$\displaystyle \sin x-\sin a=\frac{\sin(x+a)\sin(x-a)}{\sin x+\sin a}$
Similarly, $\displaystyle\cos x-\cos a=\frac{\cos^2x-\cos^2a}{\cos x+\cos a}=\frac{1-\sin^2x-(1-\sin^2a)}{\cos x+\cos a}=-\frac{\sin(x+a)\sin(x-a)}{\cos x+\cos a}$
$$\implies\lim_{x\to a}\frac{\sin a-\sin x}{\cos x-\cos a}=\cdots=\lim_{x\to a}\frac{\cos x+\cos a}{\sin x+\sin a}=\cdots$$
Here $\displaystyle a=\frac\pi6$
This method can be safely employed 
in  How to evaluate this trigonometric limit?
and in Find the limit as $x$ tends towards $\frac{\pi}{4}$
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