Find $\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$

Question

Find $$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$$ without L'Hôpital's rule.

My work:

1) I know that $$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$

2) Let $x=t-\frac{\pi}{6}$. Then

$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{3t}{\sin3t}\cdot\frac{1-2\sin \left(t-\frac{\pi}6\right)}{3t}$$

Answer 1

Let $x=t+\frac{\pi}{6}$ with $t\to 0$

$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{1-2\sin \left(t+\frac{\pi}{6}\right)}{\cos\left(3t+\frac{\pi}{2}\right)}=\lim_{t\rightarrow 0}\frac{1-\sqrt 3\sin t-\cos t}{-\sin 3t}=\frac{\sqrt 3}{3}$$

indeed

$$\frac{1-\sqrt 3\sin t-\cos t}{-\sin 3t}=\frac{3t}{-\sin 3t}\frac{1-\sqrt 3\sin t-\cos t}{3t}=\\ =-\frac{3t}{\sin 3t}\left(-\frac{\sqrt 3}{3}\frac{\sin t}{t}+\frac{1-\cos t}{3t}\right) \to-1\cdot\left(-\frac{\sqrt 3}{3}+0 \right) =\frac{\sqrt 3}{3}$$

Answer 2

Hint: Multiply numerator and denominator by $$1+2\sin(x)$$ you can write your term in the form $$\frac{\csc \left(\frac{\pi }{4}-\frac{x}{2}\right) \csc \left(\frac{x}{2}+\frac{\pi }{4}\right)}{2 (2 \sin (x)+1)}$$

Answer 3

It is helpful to shift the variable with $t=x+\frac\pi6$ (with $t\to0$), and transform to

$$\frac{1-2\sin(t+\frac\pi6)}{\cos(3t+\frac\pi2)}=\frac{1-\sqrt3\sin t-\cos t}{-\sin3t}.$$

Then $1-\cos t$ yields a quadratic term ($\sin^2\frac t2$) which can be neglected and the limit is

$$\frac{-\sqrt 3}{-3}.$$

Answer 4

Hint:

$1-2\sin(t-\frac{\pi}{6})=1-2(\sin t\cos\frac{\pi}{6}+\cos t\sin\frac{\pi}{6})=2\sin^2\frac{t}{2}-\sqrt{3}\sin t$.

Answer 5

Let $f(x) = 2 \sin x, g(x) = \cos 3x.$ Then the expression equals

$$-\frac{f(x) - f(\pi/6)}{g(x)- g(\pi/6)}= -\frac{(f(x) - f(\pi/6))/(x-\pi/6)}{(g(x)- g(\pi/6))/(x-\pi/6)}.$$

As $x\to\pi/6,$ the numerator in the fraction $\to f'(\pi/6),$ while the denominator in the fraction $\to g'(\pi/6).$ From there the limit is easy (and no, we didn't use L'Hopital).