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 $1$.

You are given 
$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$
I guess you know 
$$\lim_{x\to 0}\dfrac{\sin x}{x}=1$$
$$\lim_{x\to 0} \dfrac{e^{2x} - 1}{2x}=1$$
The most healthy way of solving this is using
$$\frac{\sin x}{x} = 1-\frac {x^2}{6}+o(x^2)$$
$$\frac{e^x-1}{x}=1+\frac x 2 +o(x^2)$$
$$\cos x = 1-\frac {x^2}{2}+o(x^2)$$
This gives
$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$
$$\eqalign{
  & \mathop {\lim }\limits_{x \to 0} \;\frac{{1 - \dfrac{{{x^2}}}{6} + o({x^2}) - 1 + \dfrac{{{x^2}}}{2} - o({x^2})}}{{2x\left( {1 + x + o({x^2}) - 1} \right)}}  \cr 
  & \mathop {\lim }\limits_{x \to 0} \;\frac{{\dfrac{{{x^2}}}{3} + o\left( {{x^2}} \right)}}{{2{x^2} + 2xo\left( {{x^2}} \right)}}  \cr 
  & \mathop {\lim }\limits_{x \to 0} \;\frac{{\dfrac{1}{3} + \dfrac{{o\left( {{x^2}} \right)}}{{{x^2}}}}}{{2 + 2\dfrac{{o\left( {{x^2}} \right)}}{{{x^2}}}}} = \dfrac{1}{6} \cr} $$
Note that
$$\eqalign{
  & \frac{{o\left( {{x^2}} \right)}}{{{x^2}}} \to 0  \cr 
  & \frac{{2o\left( {{x^2}} \right)}}{x} \to 0  \cr} $$
