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} $