Evaluating $\lim\limits_{x \to \pi/ 6} \frac{(2\sin x + \cos(6x))^2}{(6x - \pi)\sin(6x)}$ 
$$\lim_{x \to \pi/ 6} \frac{(2\sin x + \cos(6x))^2}{(6x - \pi)\sin(6x)}$$

I would expand with Maclaurin series but $x \to \frac \pi 6$ so I cannot do that. So I evaluated it with l'Hopital rule (result is $-\frac 1{12}$), but is there a better way? I had to differentiate two times and it gets really big and complicated.
 A: You can shift the argument with $x=y+\frac\pi6$:
$$\lim_{y\to0}\frac{(2\sin(y+\frac\pi6)+\cos(6y+\pi))^2}{6y\sin(6y+\pi)}=\lim_{y\to0}\frac{(\sqrt3\sin(y)+\cos(y)-\cos(6y))^2}{-6y\sin(6y)}.$$
Now it is enough to see that Taylor expansion will give $(\sqrt3y+o(y))^2=3y^2+o(y^2)$ in the numerator and $-36y^2+o(y^2)$ in the denominator, and the limit is
$$-\frac3{36}.$$
A: An elementary way is the following.
Consider the functions of real variable $f,g$ defined by $f(x)=\sin(6x)$ and $g(x)=2\sin(x)+\cos(6x)$, for all $x\in \mathbb R$. Note that $f(\pi /6)=0=g(\pi /6)$.
Now
$$\begin{align} 
\lim \limits_{x\to \pi /6}\left(\frac{(2\sin x + \cos(6x))^2}{(6x - \pi)\sin(6x)}\right)&=\lim \limits_{x\to \pi /6}\left[\left(\dfrac{2\sin(x)+\cos(6x)}{6x-\pi}\right)^2\dfrac{6x-\pi}{\sin(6x)}\right]\\
&=\left[\lim \limits_{x\to \pi /6}\left(\dfrac{2\sin(x)+\cos(6x)}{6x-\pi}\right)\right]^2\left[\lim \limits_{x\to \pi /6}\left(\dfrac{\sin(6x)}{6x-\pi}\right)\right]^{-1}\\
&=\dfrac 1 {6^2}\left[\lim \limits_{x\to \pi /6}\left(\dfrac{2\sin(x)+\cos(6x)}{x-\frac \pi 6}\right)\right]^2\left(\dfrac 1 6 \right)^{-1}\left[\lim \limits_{x\to \pi /6}\left(\dfrac{\sin(6x)}{x-\frac \pi 6}\right)\right]^{-1}\\
&=\dfrac 1 6 \left[\lim \limits_{x\to  \pi/ 6}\left(\dfrac{g(x)-g(\pi /6)}{x-\frac \pi 6}\right)\right]^2\left[\lim \limits_{x\to \pi /6}\left(\dfrac{f(x)-f(\pi /6)}{x-\frac \pi 6}\right)\right]^{-1}\\
&=\dfrac 1 6\left(g'(\pi/ 6)\right)^2\left(f'(\pi /6)\right)^{-1}\\
&=\dfrac 1 6(2\cos(\pi /6)-6\sin(\pi))^2(6\cos(\pi))^{-1}\\
&=\dfrac 1 6\sqrt 3^2\dfrac 1 6 (-1)\\
&=-\dfrac 3 {36}\\
&=-\dfrac 1{12}.
\end{align}$$
A: Write the expression as
$$\left( \frac{2\sin x+\cos 6x}{6x-\pi} \right)^2 \left(\frac{6x -\pi}{\sin 6x}\right)$$
now by L'hopital,
$$\frac{2\sin x+\cos 6x}{6x-\pi} \rightarrow 
\frac{1}{2\sqrt{3}}$$
$$\frac{6x -\pi}{\sin 6x}\rightarrow -1$$
putting it together you get $-\frac{1}{12}$
A: 1) To simplify, set $6x - \pi = v$,
2) use the angle formulas, i.e. write $\sin x = \sin (\pi -x)$ and $\cos x =-\cos(\pi-x)$. 
The first term that easily comes out after multiplying by $v$ is $-\frac{v}{ \sin v} = 1$. Can you do the rest without Taylor series expansion/L'Hospital's? 
