# 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!

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## 6 Answers

$$F=\frac{2\sin{(x)}-1}{\sqrt{3}\tan{(x)}-1} =\frac2{\sqrt3}\cdot\frac{\sin x-\sin \frac\pi6}{\tan x-\tan\frac\pi6} =\frac2{\sqrt3}\cdot\frac{2\sin \left(\frac{x-\frac\pi6}2\right)\cos \left(\frac{x+\frac\pi6}2\right)}{\tan x-\tan\frac\pi6}$$ (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, $$\lim_{x\to\frac\pi6}F=\lim_{x\to\frac\pi6}\frac2{\sqrt3} \cos x\cos\frac\pi6\frac{\cos\left(\frac{x+\frac\pi6}2\right)}{\cos\left(\frac{x-\frac\pi6}2\right)}=\frac2{\sqrt3} \cos \frac\pi6\cos\frac\pi6 \frac{\cos\frac\pi6}{\cos0}=\frac34$$

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Thanks, very nice solution. How did you come up with this tho? How did you know that this complex sequence of steps would lead you to the answer? Did it take you a long time? – Ovi May 7 '13 at 22:58
@Ovi, I've observed the Singularity at $x=\frac\pi6\implies \sin \left(\frac{x-\frac\pi6}2\right)=0$ there. Please find my other answer.Hope that is simpler. – lab bhattacharjee May 8 '13 at 4:53

Hint: $\sin(2x)=2\sin x \cos x$

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Haha more hints please? – Ovi May 6 '13 at 23:44
I don't see how this really helps – Ethan May 6 '13 at 23:47
Yea the only thing you can do after that is factor out the cos(x) but I don't see how that helps – Ovi May 6 '13 at 23:47

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.

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Is there really no other way? – Ovi May 6 '13 at 23:54
Well, this is one of the simpler methods. I suppose you could try reorganizing the limit... – Marvon May 6 '13 at 23:57
Well yea I'm trying to do that but I'm kindda stuck. – Ovi May 7 '13 at 0:06
Which part? Do you need some help with the differentiation steps for the top and bottom parts? – Marvon May 7 '13 at 0:20
No I meant I was trying to reorganize the limit and solve it without L'Hopital's rule – Ovi May 7 '13 at 0:22

Note that $$\dfrac{2 \sin(x)-1}{\sqrt3 \tan(x) - 1} = \dfrac2{\sqrt3} \left(\dfrac{\sin(x)-1/2}{\tan(x)-1/\sqrt{3}} \right) = \dfrac2{\sqrt3} \dfrac{\dfrac{\sin(x)-\sin(\pi/6)}{x-\pi/6}}{\dfrac{\tan(x)-\tan(\pi/6)}{x-\pi/6}}$$

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Oh thanks I see so now it's just the derivative of sin x at pi/6 over the derivative of the tangent. Thanks – Ovi May 7 '13 at 0:20
Wait I thought that it would just be $cos(pi/6)/sec^2(pi/6)$. However, this gives you 3sqrt(3)/8, and the correct answer is 3/4. – Ovi May 7 '13 at 0:27
@Ovi There is a $\dfrac2{\sqrt3}$ multiplying it. – user17762 May 7 '13 at 0:31
Haha ok thanks I completely missed that – Ovi May 7 '13 at 0:31

Hint: L'Hopital's Rule right at the beginning

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Is there really no other way? – Ovi May 6 '13 at 23:54

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 $y\to0,\sin y\to 0\implies \sin y\ne0$

$$=\frac{\sqrt3}4\cdot(\sqrt3+0)$$

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@Ovi, is this simpler than my other answer? – lab bhattacharjee May 11 '13 at 5:10