<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] Another way is to use change of variables. Let $y = x - \pi/6$.
$$
\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3} = \lim_{y \to 0}\frac{1 - 2\sin{(y + \pi/6)}}{2\sqrt{3}\cos{(y+\pi/6)} - 3} = L
$$
But, 
$$
1 - 2\sin(y + \pi/6) = (1 - \cos y) - \sqrt{3}\sin y
$$
and
$$
2\sqrt{3}\cos{(y+\pi/6)} - 3 = 2\sqrt{3}\bigg(\frac{\sqrt{3}}{2}\cos y - \frac{1}{2}\sin y\biggr) - 3 
$$
$$
= 3\cos y - \sqrt{3}\sin y - 3 = -3(1 - \cos y) - \sqrt{3}\sin y
$$
Thus,
$$
L = \lim_{y \to 0}\dfrac{\dfrac{1 - \cos y}{y} - \dfrac{\sqrt{3}\sin y}{y}}{-3\dfrac{1 - \cos y}{y} - \sqrt{3}\dfrac{\sin y}{y}} = 1
$$
I used the fact that
$$
\lim_{y \to 0}\frac{1 - \cos y}{y} = 0 \quad \text{and} \quad \lim_{y \to 0}\frac{\sin y}{y} = 1
$$
</s>