hmm i dont think so is there any other way than that?
Yes. I worked this out. You might approach it with the following method.
Hints:
Rewrite it as:
\(\displaystyle \displaystyle\lim_{x \to \tfrac{\pi}{6}}\bigg(\dfrac{2cos(x) \ - \ \sqrt{3}}{6(x - \tfrac{\pi}{6})}\bigg)\)
Let \(\displaystyle \ y \ = \ x - \tfrac{\pi}{6}\)
As x approaches pi/6, y approaches 0.
Substitute in the y expression:
\(\displaystyle \displaystyle\lim_{y \to 0}\bigg(\dfrac{2cos(y + \tfrac{\pi}{6}) \ - \ \sqrt{3}}{6y}\bigg)\)
Use cos(A + B), expand, and simplify.
And for this problem, assume that you have prior knowledge that
\(\displaystyle \displaystyle\lim_{y \to 0}\dfrac{cos(y) - 1}{y} \ = \ 0 \ \ \ \ and \ \ \ \ \displaystyle\lim_{y \to 0}\dfrac{sin(y)}{y} \ = \ 1\)
You must complete the supporting steps in between the hints, and following the last hints.
