<s>[INST] Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$
This is
$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
Split that:
$$\frac{\sin x\cdot \cos x}{x^2\cdot\sin 2x\cdot \cos x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
In the left side, we can cancel the $\cos x$ and also apply $\frac{\sin x}{x} = 1$ once:
$$\frac{1}{x\cdot\sin 2x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
That was probably a bad idea, since $x \cdot \sin2x$ will definitely be $0$... But anyway, let's keep going with the right side. There, we can apply the identity $\frac{\sin x}{x} = 1$ again:
$$\frac{1}{x\cdot\sin 2x} - \frac{1}{x\cdot\sin 2x\cdot \cos x}$$
Hey, I could get rid of the $\sin 2x$ on the left side if I multiply and divide by $2x$... the same on the right side:
$$\frac{1}{2x^2} - \frac{1}{2x^2\cdot \cos x}$$
Looking pretty, but sadly that's not going anywhere. What can I do?
 [/INST] Notice, $$\lim_{x\to 0}\frac{\sin x-\tan x}{x^2\sin 2x}$$
$$=\lim_{x\to 0}\frac{\sin x\frac{(\cos x-1)}{\cos x}}{2x^2\sin x\cos x}$$
$$=\frac{1}{2}\lim_{x\to 0}\frac{\cos x-1}{x^2\cos^2 x}$$
$$=\frac{1}{2}\lim_{x\to 0}\frac{\cos x-1}{x^2}\cdot \lim_{x\to 0}\frac{1}{\cos^2x}$$
$$=\frac{1}{2}\lim_{x\to 0}\frac{\left(1-\frac{x^2}{2!}+O(x^2)\right)-1}{x^2}\cdot 1$$
$$=\frac{1}{2}\lim_{x\to 0}\frac{\left(-\frac{x^2}{2!}+O(x^2)\right)}{x^2}$$
$$=\frac{1}{2}\lim_{x\to 0}\left(-\frac{1}{2!}+O(1)\right)$$
$$=\frac{1}{2}\left(-\frac{1}{2}+0\right)=\color{red}{-\frac{1}{4}}$$
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