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I want to compute this limit:

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

Using L'Hopital, is easy to get the result, which is $\sqrt{3}$

I tried using linear approximation (making $u = x - \displaystyle \frac{\pi}{3}$)

$\displaystyle \lim_{u\to 0} \frac{1 - 2\cos(u + \frac{\pi}{3})}{\sin(u)} = \lim_{u\to 0} \frac{1 - \cos(u) + \sqrt{3}\sin(u)}{u} \approx \lim_{u\to 0} \frac{1 - 1 + \sqrt{3}u}{u} = \sqrt{3}$

But it bothers me using that sort of linear approximation, I want to get the result in a more formal way. I have tried using double angle properties

$$\cos(2x) = \cos^2(x) - \sin^2(x)$$ $$\sin(2x) = 2\sin(x)\cos(x)$$

But I reach to a point of nowhere, I cannot come up with a way of simplifying expressions to get the results:

$\displaystyle\lim_{x\to\pi/3} 2\frac{3\sin^2(\frac{x}{2}) - \cos^2(\frac{x}{2})}{\sqrt{3}\sin(x) - \cos(x)}$

Is there a way of computing this limit without approximations and without L'Hopital?

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  • $\begingroup$ Are you willing to accept $\lim_{u\to0}\frac{\sin u}{u}=1$? $\endgroup$ Apr 25, 2017 at 5:46
  • $\begingroup$ a linear approximation is L'hopital right before u take the derivative $\endgroup$ Apr 25, 2017 at 5:48
  • $\begingroup$ that approach using linear approximations is wrong. limits are not approximations contrary to what many beginners in calculus think. $\endgroup$
    – Paramanand Singh
    Apr 25, 2017 at 6:30

5 Answers 5

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HINT:

Use $1-\cos(u)=2\sin^2(u/2)$ along with

$$\lim_{u\to 0}\frac{\sin(u)}{u}=1$$

and

$$\lim_{u\to 0}\frac{\sin^2(u/2)}{u}=0$$

Note that in the OP, $1-2\cos(x)=1-\cos(u)\color{blue}{+}\sqrt 3 \sin(u)$

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HINT:

Using Prosthaphaeresis & Double angle Formula as $\cos\dfrac\pi3=?$

$$2\cdot\dfrac{\cos\dfrac\pi3-\cos x}{\sin\left(x-\dfrac\pi3\right)}=2\cdot\dfrac{2\sin\dfrac{\pi+3x}6\sin\dfrac{3x-\pi}6}{2\sin\dfrac{3x-\pi}6\cos\dfrac{3x-\pi}6}$$

Now as $x\to\dfrac\pi3\iff3x\to\pi,3x\ne\pi\implies\sin\dfrac{3x-\pi}6\ne0$, so it can be cancelled safely.

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  • $\begingroup$ More generally, $$\dfrac{\cos2A-\cos2y}{\sin(2y-2A)}=\dfrac{2\sin(y+a)\sin(y-a)}{2\sin(y-a)\cos(y-a)}$$ Here $2A=\dfrac\pi3,2y=x$ $\endgroup$ Apr 25, 2017 at 9:25
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Note that $$\frac{\sin u}{u}\to 1$$ and $$\frac{1-\cos(u)}{u}=u\frac{1-cos u}{u^2}\to 0$$

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Note that $\cos (\pi/3) = 1/2$ and hence we can write $$\lim_{x \to \pi/3}\frac{1 - 2\cos x}{\sin (x - \pi/3)} = -2\lim_{x \to \pi/3}\frac{\cos x - \cos (\pi/3)}{x - \pi/3}\cdot\frac{x - \pi/3}{\sin (x - \pi/3)} = -2(-\sin \pi/3)\cdot 1 = \sqrt{3}$$ here we have used the fact that $(\cos x)' = -\sin x$ so that $$\lim_{x \to a}\frac{\cos x - \cos a}{x - a} = -\sin a$$

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This is actually much easier than it looks. By letting $x=u+\frac\pi3$, we get

$$ \lim_{x\to\pi/3} \frac{1-2\cos(x)}{\sin(x-\frac\pi3)} = \lim_{u\to0}\frac{1-\cos(u)+\sqrt{3}\sin(u)}{\sin(u)} $$ Now, we write the limit on the right as $$ \lim_{u\to0} \frac{1-\cos(u)}{\sin(u)} + \lim_{u\to0}\frac{\sqrt{3}\sin(u)}{\sin(u)}=\lim_{u\to0} \frac{1-\cos(u)}{\sin(u)} + \sqrt{3} $$ Multiplying the remaining limit by $\frac{1+\cos(u)}{1+\cos(u)}$ gives $$\begin{align} \lim_{u\to0} \frac{1-\cos^2(u)}{\sin(u)(1+\cos(u))} + \sqrt{3}&=\lim_{u\to0} \frac{\sin^2(u)}{\sin(u)(1+\cos(u))} + \sqrt{3}\\ &=\lim_{u\to0} \frac{\sin(u)}{1+\cos(u)} + \sqrt{3}\\ &=\sqrt{3} \end{align}$$

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