# Estimate limit without using L'Hopital [closed]

It is possible to estimate the following limit without using L'Hopital Rule?

$$\lim_{x\to \frac{\pi }{3}}\left(\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}\right)$$

I will be happy to get any tip!

• You can try using taylor series – Libertas Mar 6 at 22:58
• One of my professors at the university was very against L'Hopital, because it can be replaced by using enough taylor expansions which are often easier to use (and in general have many more uses and should thus be taught more). – Bakuriu Mar 7 at 8:54

• First, recenter around $$0$$, because $$0$$ is nicer to think about, and it's easier to recognize identities around it: $$\lim_{x\to \frac{\pi }{3}} \frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos x} = \lim_{x\to 0} \frac{\sin x}{1-2\cos\left(x+\frac{\pi }{3}\right)}$$

• Next, using the identity $$\cos(a+b)=\cos a \cos b-\sin a \sin b$$, you get $$1-2\cos\left(x+\frac{\pi }{3}\right) = 1-\cos x + \sqrt{3}\sin x$$

• At this point, you may "expect" $$\sin x \approx x$$ and $$1-\cos x \approx x^2/2$$ around $$0$$ (by a Taylor series expansion), which tells you the limit should be $$1/\sqrt{3}$$. If you don't have that intuition, no problem! It's not necessary. What we want to use are standard limits around $$0$$, like $$\lim_{x\to 0}\frac{\sin x}{x} = 1$$.

• You can rewrite $$\frac{\sin x}{1-2\cos\left(x+\frac{\pi }{3}\right)} = \frac{\sin x}{1-\cos x + \sqrt{3}\sin x} = \frac{1}{\frac{1-\cos x}{x}\cdot \frac{x}{{\sin x}} + \sqrt{3}}$$ and use the facts that $$\lim_{x\to0} \frac{1-\cos x}{x} = 0, \qquad \lim_{x\to0} \frac{\sin x}{x} = 1$$ (can you see why?) to conclude that $$\lim_{x\to\frac{\pi}{3}}\frac{\sin x}{1-2\cos\left(x+\frac{\pi }{3}\right)} = \frac{1}{0\cdot 1 + \sqrt{3}} = \frac{1}{\sqrt{3}}$$

This is a variant on Clement C.'s answer, showing a way to further simplify the calculation of the limit.

After recentering the limit around $$x\to0$$ and showing that $$1-2\cos(x-\pi/3)=1-\cos x+\sqrt3\sin x$$, multiply numerator and denominator by $$1+\cos x$$, so that we have

\begin{align} {\sin x\over1-2\cos(x-{\pi\over3})} &={\sin x\over1-\cos x+\sqrt3\sin x}\cdot{1+\cos x\over1+\cos x}\\ &={\sin x(1+\cos x)\over1-\cos^2x+\sqrt3\sin x(1+\cos x)}\\ &={\sin x(1+\cos x)\over\sin^2x+\sqrt3\sin x(1+\cos x)}\\ &={1+\cos x\over\sin x+\sqrt3(1+\cos x)}\\ &\to{1+1\over0+\sqrt3(1+1)}={1\over\sqrt3}\\ \end{align}