# Calculus - limit of a function: $\lim\limits_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$

How do you compute the following limit without using the l'Hopital rule? If you were allowed to use it, it becomes easy and the result is $\sqrt{3}\over 3$ but without it, I am not sure how to proceed. $$\lim_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$$

• i would set $t=x-\frac{\pi}{3}$ – Dr. Sonnhard Graubner Dec 14 '15 at 18:05
• I guess you can use the Taylor expansions of sine and cosine. – This Is Me Dec 14 '15 at 18:05
• Unnecessary, if you substitute and use the formula for cos(x+y), everything cancels. – orion Dec 14 '15 at 18:06
• Can't we use L'Hopital Rule? – user297008 Dec 14 '15 at 18:34

Let $y=x-\pi/3$. Then, the limit of interest becomes

\begin{align} \lim_{y\to 0}\frac{\sin y}{1-2\cos (y+\pi/3)}&=\lim_{y\to 0}\frac{\sin y}{1-\cos y+\sqrt 3 \sin y}\\\\ &=\lim_{y\to 0}\frac{2\sin (y/2)\cos(y/2)}{2\sin^2(y/2)+2\sqrt{3}\sin(y/2)\cos(y/2)}\\\\ &=\lim_{y\to 0}\frac{\cos(y/2)}{\sin(y/2)+\sqrt 3 \cos(y/2)}\\\\ &=\frac{\sqrt 3}{3} \end{align}

$$\frac {\sin x\cos \frac {\pi}{3}- \cos x\sin\frac {\pi}{3}}{1+2\cos x}\\ \frac {\frac 12\sin x- \frac {\sqrt 3}{2} \cos x}{1-2\cos x}\\ \frac {\sin x- \sqrt 3\cos x}{2-4\cos x}\\ \frac {\sin^2 x- 3\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\ \frac {\sin^2 x + \cos^2 x- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\ \frac {1- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\ \frac {(1- 2\cos x)(1+2\cos x)}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\ \frac {(1+2\cos x)}{2(\sin x + \sqrt 3 \cos x)}$$

And now we can plug $$x = \frac {\pi}{3}$$

$$\frac {2}{2(\frac {\sqrt 3}{2} + \frac {\sqrt 3}{2})}$$

$$\frac {1}{\sqrt 3}$$

Proceed as follows:

$$\lim_{x\to\frac{\pi}{3}} \frac{\sin(x-\pi/3)}{1-2\cos x}$$ $$=\lim_{t\to0} \frac{\sin t}{1-2\cos (t+\frac{\pi}{3})}$$ $$=\lim_{t\to0} \frac{\sin t}{1-2\cos t\cos \frac{\pi}{3}+2\sin t \sin\frac{\pi}{3}}$$ $$=\lim_{t\to0} \frac{\sin t}{(1-\cos t)+2\sin t \sin\frac{\pi}{3}}=\frac{1}{2\sin\frac{\pi}{3}}=\frac{1}{\sqrt3}$$

Maybe a little bit of reasoning: $1-\cos t\approx t^2/2$ for small $t$, which is insignificant compared to $\sin t$. If unsure, maybe use $1-\cos t = 2\sin^2 \frac{t}{2}$ and put everything else into half angles to. In that case, you get

$$=\lim_{t\to0} \frac{2\sin \frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}+4\sin\frac{t}{2}\cos\frac{t}{2} \sin\frac{\pi}{3}}=\lim_{t\to0} \frac{2\cos\frac{t}{2}}{2\sin\frac{t}{2}+4\cos\frac{t}{2} \sin\frac{\pi}{3}}=\frac{2}{4\sin\frac{\pi}{3}}=\frac{1}{\sqrt3}$$