Find $$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$$ without L'Hôpital's rule.
1) I know that $$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$
2) Let $x=t-\frac{\pi}{6}$. Then
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{3t}{\sin3t}\cdot\frac{1-2\sin \left(t-\frac{\pi}6\right)}{3t}$$
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).
Hint: Multiply numerator and denominator by $$1+2\sin(x)$$ you can write your term in the form $$\frac{\csc \left(\frac{\pi }{4}-\frac{x}{2}\right) \csc \left(\frac{x}{2}+\frac{\pi }{4}\right)}{2 (2 \sin (x)+1)}$$
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
$$\frac{-\sqrt 3}{-3}.$$
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
$$\frac{1-\sqrt 3\sin t-\cos t}{-\sin 3t}=\frac{3t}{-\sin 3t}\frac{1-\sqrt 3\sin t-\cos t}{3t}=\\ =-\frac{3t}{\sin 3t}\left(-\frac{\sqrt 3}{3}\frac{\sin t}{t}+\frac{1-\cos t}{3t}\right) \to-1\cdot\left(-\frac{\sqrt 3}{3}+0 \right) =\frac{\sqrt 3}{3}$$
Hint:
$1-2\sin(t-\frac{\pi}{6})=1-2(\sin t\cos\frac{\pi}{6}+\cos t\sin\frac{\pi}{6})=2\sin^2\frac{t}{2}-\sqrt{3}\sin t$.
