Limit involving $(\sin x) /x -\cos x $ and $(e^{2x}-1)/(2x)$, without l'Hôpital Find:
$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$
I have factorized it in this manner in an attempt to use the formulae.
I have tried to use that for $x$ tending to $0$, $\dfrac{\sin x}{x} = 1$ and that $\dfrac{e^x - 1}x$ is also $1$.

$$\dfrac{\sin(x)}{x} = 1 - \dfrac{x^2}{3!} + \mathcal{O}(x^4)$$
$$\cos(x) = 1 - \dfrac{x^2}{2!} + \mathcal{O}(x^4)$$
$$\exp(2x) = 1 + 2x + \dfrac{(2x)^2}{2!} + \mathcal{O}(x^3)$$
Hence, $$\dfrac{\dfrac{\sin(x)}{x} - \cos(x)}{\exp(2x) - 1 -2x} = \dfrac{-\dfrac{x^2}{3!} + \dfrac{x^2}{2!} + \mathcal{O}(x^4)}{2x^2 + \mathcal{O}(x^3)} = \dfrac{\dfrac{x^2}{3} + \mathcal{O}(x^4)}{2x^2 + \mathcal{O}(x^3)} = \dfrac{\dfrac13 + \mathcal{O}(x^2)}{2 + \mathcal{O}(x)}$$
Hence, $$\lim_{x \rightarrow 0} \dfrac{\dfrac{\sin(x)}{x} - \cos(x)}{\exp(2x) - 1 -2x} = \dfrac16$$
