hi, i need urgent to solve this limit $\displaystyle \lim_{x \to \pi/6} \frac{2cosx-\sqrt3}{sin(\pi/6-x)}$

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- Oct 25th 2008, 09:06 AMironman2trig limit
hi, i need urgent to solve this limit $\displaystyle \lim_{x \to \pi/6} \frac{2cosx-\sqrt3}{sin(\pi/6-x)}$

- Oct 25th 2008, 09:07 AMKrizalid
Put $\displaystyle z=\frac\pi6-x,$ remaining stuff is just a matter of algebra.

- Oct 25th 2008, 09:19 AMironman2
can you do the whole solution please?

- Oct 25th 2008, 09:25 AMPeritus
L'Hospital

- Oct 25th 2008, 09:29 AMSoroban
Hello, ironman2!

Are you allowed to use L'Hopital?

Quote:

$\displaystyle \lim_{x \to\frac{\pi}{6}} \frac{2\cos x-\sqrt3}{\sin(\frac{\pi}{6}-x)}$

Apply L'Hopital: .$\displaystyle \lim_{x\to\frac{\pi}{6}}\left[ \frac{-2\sin x}{-\cos(\frac{\pi}{6}-x)}\right] \;=\; \lim_{x\to\frac{\pi}{6}}\frac{2\sin x}{\cos(\frac{\pi}{6}-x)}$

. . $\displaystyle =\;\frac{2\sin\frac{\pi}{6}}{\cos(\frac{\pi}{6}-\frac{\pi}{6})} \;=\;\frac{2\cdot\frac{1}{2}}{\cos 0} \;=\;\frac{1}{1} \;=\;1$

- Oct 25th 2008, 09:39 AMKrizalid
Put $\displaystyle z=\frac\pi6-x$ and the limit becomes,

$\displaystyle \underset{z\to 0}{\mathop{\lim }}\,\frac{\sqrt{3}(\cos z-1)+\sin z}{\sin z}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\dfrac{\sqrt{3}(\cos z-1)}{z}+\dfrac{\sin z}{z}}{\dfrac{\sin z}{z}}=1.$ - Oct 25th 2008, 01:04 PMironman2
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. - Oct 25th 2008, 01:33 PMmr fantastic