limits - Evaluating $limlimits_{x to pi/ 6} frac{(2sin x + cos(6x))^2}{(6x - pi)sin(6x)}$

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

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.

Answer

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

$$-\frac3{36}.$$