Showing $ \lim_{x \to \pi/3} \frac{1-2\cos x}{\sin\left(x-\frac{\pi}{3}\right)}=\sqrt{3} $ without l'Hopital

Question

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

Getting the answer $\sqrt{3}$ is easy with l'Hopital but I want to try to do this in a more formal way. I got to

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

but im not sure what to do with $1-2\cos x$. The problem is from N. Piskunov Differential and Integral Calculus.

Answer 1

Instead of working on the denominator, it is much more effective to first let $y = x - \frac{\pi}{3}$ so that $$L = \lim_{x \to \pi/3} \frac{1 - 2 \cos x}{\sin (x - \frac{\pi}{3})} = \lim_{y \to 0} \frac{1 - 2 \cos (y + \frac{\pi}{3})}{\sin y}.$$

Now use trigonometric identities on the numerator: $$1 - 2 \cos (y + \tfrac{\pi}{3}) = 1 - \cos y + \sqrt{3} \sin y$$ and now we see $$L = \sqrt{3} + \lim_{y \to 0} \frac{1 - \cos y}{\sin y}.$$ Finally, apply one more trigonometric identity: $$\frac{1 - \cos y}{\sin y} = \frac{1 - (1 - 2 \sin^2 \frac{y}{2})}{2 \sin \frac{y}{2} \cos \frac{y}{2}} = \tan \frac{y}{2},$$ so $$L = \sqrt{3}.$$

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Alternatively, we could have also written $$\frac{1 - \cos y}{\sin y} = \frac{(1 - \cos y)(1 + \cos y)}{\sin y (1 + \cos y)} = \frac{1 - \cos^2 y}{\sin y (1 + \cos y)} = \frac{\sin y}{1 + \cos y}$$ and let $y \to 0$.

Answer 2

Convert the trig limit into a polynomial, by using $t = \tan{x/2}$. It is somewhat extensive but polynomial limits are easier to evaluate.

$t = \tan{x/2} \implies \sin{x} = \dfrac{2t}{1 + t^2}, \, \cos{x} = \dfrac{1 - t^2}{1 + t^2}$

Hence,

$$\lim\limits_{x \to \pi/3}{\frac{2(1 - 2\cos{x})}{\sin{x} - \sqrt{3}\cos{x}}} \xrightarrow{t \, = \, \tan{x/2}} \lim\limits_{t \to \sqrt{3}/3}{\frac{2 - 4\left(\frac{1 - t^2}{1 + t^2}\right)}{\left(\frac{2t}{1 + t^2}\right) - \sqrt{3}\left(\frac{1 - t^2}{1 + t^2}\right)}} \\=\lim\limits_{t \to \sqrt{3}/3}{\frac{\left(\frac{2 + 2t^2 + 4t^2 - 4}{1 + t^2}\right)}{\left(\frac{\sqrt{3}t^2 + 2t - \sqrt{3}}{1 + t^2}\right)}} \\=\lim\limits_{t \to \sqrt{3}/3}{\frac{2(3t^2 - 1)}{\sqrt{3}t^2 + 2t - \sqrt{3}}}$$

See that it is a $0/0$ situation at the limit, so $t = \sqrt{3}/3$ is a root of both polynomials. So we factor, $(3t^2 - 1) = (\sqrt{3}t - 1)(\sqrt{3}t + 1)$

and

$\sqrt{3}t^2 + 2t - \sqrt{3} = (t - \sqrt{3}/3)(\sqrt{3}t + 3) = (\sqrt{3}/3)(\sqrt{3}t - 1)(\sqrt{3}t + 3)$

$$\lim\limits_{t \to \sqrt{3}/3}{\frac{2(\sqrt{3}t - 1)(\sqrt{3}t + 1)}{(\sqrt{3}/3)(\sqrt{3}t - 1)(\sqrt{3}t + 3)}} \\=\lim\limits_{t \to \sqrt{3}/3}{\frac{2(\sqrt{3}t + 1)}{(\sqrt{3}/3)(\sqrt{3}t + 3)}} \\=\sqrt{3}$$

Answer 3

Hint: You can actually solve this question by the definition of derivative which is very useful in some special cases, but always being ignored by people.
$$ \begin{aligned} L&=\lim_{x\to\frac{\pi}{3}}\frac{1-2\cos x}{\sin(x-\frac{\pi}{3})}\\ &=\lim_{x\to\frac{\pi}{3}}\frac{x-\frac{\pi}{3}}{\sin(x-\frac{\pi}{3})}\times \lim_{x\to\frac{\pi}{3}}\frac{1-2\cos x}{x-\frac{\pi}{3}}\\ &=\lim_{x\to\frac{\pi}{3}}\frac{1-2\cos x}{x-\frac{\pi}{3}}\\ &=-2\lim_{x\to\frac{\pi}{3}}\frac{\cos x-\frac{1}{2}}{x-\frac{\pi}{3}}\\ &=-2\lim_{x\to\frac{\pi}{3}}\frac{\cos x-\cos\frac{\pi}{3}}{x-\frac{\pi}{3}}\\ &=-2\cos'x|_{x=\frac{\pi}{3}}\\ &=2\sin x|_{x=\frac{\pi}{3}}\\\\ &=\sqrt{3} \end{aligned} $$
I hope it will help.
Also, you can use the Taylor Expansion, but I think it is using the same principle with the L'$\hat H$opital rule.

Answer 4

$$1-2\cos x =\\2(\frac 12 - \cos x)\\=2(\cos \frac{\pi}{3} -\cos x)\\$$now can use sum to product formula $$\cos A - \cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$$ it means $$1-2\cos x=\\2(\cos \frac{\pi}{3} -\cos x)=\\2(-2\sin(\frac{\frac{\pi}{3}+x}{2})\sin(\frac{\frac{\pi}{3}-x}{2}))=+4\sin(\frac{\frac{\pi}{3}+x}{2})\sin(\frac{x-\frac{\pi}{3}}{2})$$ and finally $$\lim_{x \to \frac{\pi}{3}} \frac{(1-2\cos x)}{\sin(x-\frac{\pi}{3})}=\\\lim_{x \to \frac{\pi}{3}} \frac{4\sin(\frac{\frac{\pi}{3}+x}{2})\sin(\frac{x-\frac{\pi}{3}}{2})}{\sin(x-\frac{\pi}{3})}=\\\lim_{x \to \frac{\pi}{3}} \frac{\sin(\frac{x-\frac{\pi}{3}}{2})}{\sin(x-\frac{\pi}{3})}\times \lim_{x \to \frac{\pi}{3}}4\sin(\frac{\frac{\pi}{3}+x}{2})= \\\frac 12 \times 4\sin \frac{\pi}{3}$$

Answer 5

You could use double-angle and sum-to-product formulas.

$\begin{align}\!\!\!\!\!\!\lim\limits_{x\to\frac\pi3}\dfrac{1\!-\!2\cos x}{\sin\left(x\!-\!\frac\pi3\right)}\!=\!\lim\limits_{x\to\frac\pi3}\!\dfrac{-2\!\left(\cos x\!-\!\frac12\right)}{\sin\left(x\!-\!\frac\pi3\right)}\!&=\!\lim\limits_{x\to\frac\pi3}\!\dfrac{-2\!\left(\cos x\!-\!\cos\!\frac\pi3\right)}{\sin\left(x\!-\!\frac\pi3\right)}\!=\end{align}$

$=\require{cancel}\lim\limits_{x\to\frac\pi3}\dfrac{4\sin\left(\frac{x+\pi/3}2\right)\color{#8888FF}{\cancel{\color{black}{\sin\left(\frac{x-\pi/3}2\right)}}}}{2\color{#8888FF}{\cancel{\color{black}{\sin\left(\frac{x-\pi/3}2\right)}}}\!\cos\left(\frac{x-\pi/3}2\right)}=\lim\limits_ {x\to\frac\pi3}\dfrac{2\sin\left(\frac{x+\pi/3}2\right)}{\cos\left(\frac{x-\pi/3}2\right)}=$

$=\dfrac{2\sin\!\frac\pi3}{\cos0}=\sqrt3\;.$