Cool limit 2

You can solve the limit using L'Hôpital's Rule as done in @B.S.Bharath Sai Guhan 's solution but I'll provide a solution without using it. So, here we go.

We'll use these 2 identities and some of the identities listed here to solve the problem:

$1-\cos x=2\sin^2(\frac{x}{2}) \\ \cos(x\pm y)=\cos x \cos y \mp \sin x \sin y$

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Also, we will use the substitutions $y=x-\frac{\pi}{6}$ and $z=\frac{y}{2}$ in the problem and we can see that $x\to \frac{\pi}{6} \implies y\to 0 \land z\to 0$ . Now, let's start evaluating the limit.

$\displaystyle \lim_{x\to \frac{\pi}{6}} \bigg( \frac{2-\sqrt{3}\cos x - \sin x}{(6x- \pi)^2} \bigg) \\= \displaystyle 2\centerdot \lim_{x\to \frac{\pi}{6}} \bigg( \frac{1-\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x}{(6x- \pi)^2} \bigg) \\= \displaystyle 2\centerdot \lim_{x\to \frac{\pi}{6}} \bigg( \frac{1-\cos(x-\frac{\pi}{6})}{(6x- \pi)^2} \bigg) \\= \displaystyle 2\centerdot \lim_{y\to 0} \bigg( \frac{1-\cos y}{(6y)^2} \bigg) \\= \displaystyle 2\centerdot \lim_{y\to 0} \bigg( \frac{2\sin^2(\frac{y}{2})}{36y^2} \bigg) \\= \displaystyle 2\centerdot \frac{2}{36}\centerdot(\frac{1}{2})^2\centerdot \bigg(\lim_{z\to 0} \bigg( \frac{\sin z}{z} \bigg)\bigg)^2 \\= \frac{1}{36}\times 1^2 = \boxed{\frac{1}{36}}$

So, the limit is in the form of $\frac{p}{q}$ with $p=1,q=36$ and they are coprime numbers. So, the required value $=p+q=1+36=\boxed{37}$

Plug $\displaystyle\frac{\pi}{6}$ into the function. We get the function value to be indeterminate i.e $\displaystyle\frac{0}{0}$

Applying L'Hôpital's Rule, the limit becomes :

$\displaystyle\lim_{x\rightarrow \frac{\pi}{6}} \frac{\sqrt{3}sinx - cosx}{72x-12{\pi}}$

Plugging in $\displaystyle\frac{\pi}{6}$ into the function still results in $\displaystyle\frac{0}{0}$ .

Applying L'Hôpital's Rule once more, the limit becomes :

$\displaystyle\lim _{x\rightarrow \frac { \pi }{6}}{\quad\frac {\sqrt {3}\cos{x}+\sin{x}}{72}}$

Plugging in $\displaystyle\frac{\pi}{6}$ , and bringing the fraction to its lowest terms, we get $\frac{1}{36}\Rightarrow \boxed{37}$