hi, i need urgent to solve this limit $\displaystyle \lim_{x \to \pi/6} \frac{2cosx-\sqrt3}{sin(\pi/6-x)}$
Hello, ironman2!
Are you allowed to use L'Hopital?
$\displaystyle \lim_{x \to\frac{\pi}{6}} \frac{2\cos x-\sqrt3}{\sin(\frac{\pi}{6}-x)}$
Apply L'Hopital: .$\displaystyle \lim_{x\to\frac{\pi}{6}}\left[ \frac{-2\sin x}{-\cos(\frac{\pi}{6}-x)}\right] \;=\; \lim_{x\to\frac{\pi}{6}}\frac{2\sin x}{\cos(\frac{\pi}{6}-x)}$
. . $\displaystyle =\;\frac{2\sin\frac{\pi}{6}}{\cos(\frac{\pi}{6}-\frac{\pi}{6})} \;=\;\frac{2\cdot\frac{1}{2}}{\cos 0} \;=\;\frac{1}{1} \;=\;1$
Put $\displaystyle z=\frac\pi6-x$ and the limit becomes,
$\displaystyle \underset{z\to 0}{\mathop{\lim }}\,\frac{\sqrt{3}(\cos z-1)+\sin z}{\sin z}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\dfrac{\sqrt{3}(\cos z-1)}{z}+\dfrac{\sin z}{z}}{\dfrac{\sin z}{z}}=1.$
Soroban, first thank you very much. But it's one thing left, how did u changed numerator from 2cosx-sqrt3 to -2sinx? With other words how to apply this Hopital Rule (i didn't know for this since it was mentioned here). Can you explain with your words what did you do? Sorry for this ironic questions, but I'm just trying to learn, and I don't understand other literatures about this rule.
A little help would be highly appreciated.
p.s sorry for my bad english.