Solve the following limit? Please only give hints? $$\lim_{x \to \frac{\pi}{6}}\frac{2\sin{(x)}-1}{\sqrt{3}\tan{(x)}-1}$$ I tried this and I was able to simplify it down to $$\lim_{x \to \frac{\pi}{6}}\frac{\sin{(2x)}-\cos{(x)}}{2\sin{(x - \frac{\pi}{6})}}.$$ However, I am stuck here and I don't even know if this is the right approach. Thanks!
 A: Hint: $\sin(2x)=2\sin x \cos x$
A: Another hint: Try using L'Hôpital's rule. If you plug $\displaystyle \frac{\pi}{6}$ into the original limit, you'll see that it evaluates to an indeterminate $\displaystyle \frac{0}{0}$. Use that and the rule, and you should be able to solve for the limit.
A: Note that
$$\dfrac{2 \sin(x)-1}{\sqrt3 \tan(x) - 1} = \dfrac2{\sqrt3} \left(\dfrac{\sin(x)-1/2}{\tan(x)-1/\sqrt{3}} \right) = \dfrac2{\sqrt3} \dfrac{\dfrac{\sin(x)-\sin(\pi/6)}{x-\pi/6}}{\dfrac{\tan(x)-\tan(\pi/6)}{x-\pi/6}}$$
A: Hint: L'Hopital's Rule right at the beginning
A: $$F=\frac{2\sin{(x)}-1}{\sqrt{3}\tan{(x)}-1} =\frac2{\sqrt3}\cdot\frac{\sin x-\sin \frac\pi6}{\tan x-\tan\frac\pi6} =\frac2{\sqrt3}\cdot\frac{2\sin \left(\frac{x-\frac\pi6}2\right)\cos  \left(\frac{x+\frac\pi6}2\right)}{\tan x-\tan\frac\pi6}$$ (Applying $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2$)
Now, $$\tan x-\tan\frac\pi6 =\frac{\sin \left(x-\frac\pi6\right)}{\cos x\cos\frac\pi6}=\frac{2\sin \left(\frac{x-\frac\pi6}2\right)\cos\left(\frac{x-\frac\pi6}2\right)}{\cos x\cos\frac\pi6}$$
So, $$F=\frac2{\sqrt3} \cos x\cos\frac\pi6\frac{2\sin \left(\frac{x-\frac\pi6}2\right)\cos\left(\frac{x+\frac\pi6}2\right)}{2\sin \left(\frac{x-\frac\pi6}2\right)\cos\left(\frac{x-\frac\pi6}2\right)}$$
If $x\to\frac\pi6, \sin \left(\frac{x-\frac\pi6}2\right)\to0\implies \sin \left(\frac{x-\frac\pi6}2\right)\ne0$
So, $$\lim_{x\to\frac\pi6}F=\lim_{x\to\frac\pi6}\frac2{\sqrt3} \cos x\cos\frac\pi6\frac{\cos\left(\frac{x+\frac\pi6}2\right)}{\cos\left(\frac{x-\frac\pi6}2\right)}=\frac2{\sqrt3} \cos \frac\pi6\cos\frac\pi6 \frac{\cos\frac\pi6}{\cos0}=\frac34$$
A: Putting $x=\frac\pi6-y$ as $x\to\frac\pi6,y\to0$
$$2\sin x-1=2\sin(\frac\pi6-y)-1=\cos y-\sqrt3\sin y-1$$
$$\sqrt3\tan x-1=\sqrt3\tan(\frac\pi6-y)-1=\sqrt3\cdot \frac{\frac1{\sqrt3}-\tan y}{1+\tan y\frac1{\sqrt3}}-1$$
$$=\sqrt3\cdot\frac{1-\sqrt3\tan y}{\sqrt3+\tan y}-1=\frac{\sqrt3-3\tan y}{\sqrt3+\tan y}-1=\frac{-4\tan y}{\sqrt3+\tan y}$$
$$\text{So,}\frac{2\sin x-1}{\sqrt3\tan x-1}=\frac{(\cos y-\sqrt3\sin y-1)(\sqrt3+\tan y)}{-4\tan y}$$
$$=\frac14\cdot(\sqrt3+\tan y)\cdot\left(\sqrt3\frac{\sin y}{\tan y}+\frac{1-\cos y}{\tan y} \right)$$
$$\text{So,}\lim_{x\to\frac\pi6}\frac{2\sin x-1}{\sqrt3\tan x-1}$$
$$=\lim_{y\to0}\frac14\cdot(\sqrt3+\tan y)\cdot\left(\sqrt3\frac{\sin y}{\tan y}+\frac{1-\cos y}{\tan y} \right)$$
$$= \frac14\cdot\lim_{y\to0}(\sqrt3+\tan y)\cdot\left(\sqrt3 \cdot\lim_{y\to0}\frac{\sin y}{\tan y}+\lim_{y\to0}\frac{\sin^2y\cdot\cos y}{\sin y(1+\cos y)} \right)$$
$$=\frac14\cdot(\sqrt3+0)\cdot\left(\lim_{y\to0}\sqrt3\cos y+\lim_{y\to0}\frac{\sin y\cdot\cos y}{1+\cos y} \right)$$ as $y\to0,\sin y\to 0\implies \sin y\ne0$
$$=\frac{\sqrt3}4\cdot(\sqrt3+0)$$
