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?

$$\frac{\sin x-\tan x}{x^{2}\sin 2x}=\frac{\sin x}{\sin 2x}\frac{\cos x -1}{x^{2} \cos x}=\frac{\sin x}{\sin 2x}\frac{-2\sin^{2}\frac{x}{2}}{x^{2}\cos x}$$
Now, $$\frac{\sin x}{\sin 2x}\to \frac{1}{2}$$
$$\frac{\sin^{2}\frac{x}{2}}{x^{2}}=1/4\frac{\sin^{2}\frac{x}{2}}{(x/2)^{2}}\to 1/4$$
And $\cos x \to 1$
So the limit is $-1/4$.
