What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit 
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ 
for $a,b \in \rm{I\!R}_{+}$.  Applying L'Hospital's rule leads to 
$\lim_{x \rightarrow0}\frac{\cos x \cdot \sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}{-2 \sin x \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x}) + \frac{\sin x \cdot (a^2 + b^2 + 2 ab \cos x )}{\sqrt{a^2+b^2+ 2 ab \cos x}} }$.
However, this remains with both, cosine and sine.  Maybe one could use a trigonometric identity, which I cannot find. 

Note
$$\begin{align}
L = & \lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}\\
= & \lim_{x \rightarrow0} \frac{1}{2\sqrt{a^2 + b^2 + 2 ab \cos x  }}
\cdot \lim_{x \rightarrow0} \frac{a b \sin x}{\sqrt{ a+b - \sqrt{a^2+b^2+ 2 ab \cos x}}}\\
=&\frac{1}{2(a+b)}\cdot L_1 \tag1\\
\end{align}$$
where
$$\begin{align}
L_1= &  \lim_{x \rightarrow0} \frac{a b \sin x}{\sqrt{a+b - \sqrt{a^2+b^2+ 2 ab \cos x}}} \\
= &  \lim_{x \rightarrow0}  \frac{a b \sin x\sqrt{a+b + \sqrt{a^2+b^2+ 2 ab \cos x}}}{\sqrt{(a+b)^2 - (a^2+b^2+ 2 ab \cos x)}} \\
=&  \lim_{x \rightarrow0}  \frac{a b \sin x\sqrt{a+b + \sqrt{a^2+b^2+ 2 ab \cos x}}}{\sqrt{ 2ab(1-\cos x)}} \\
=&  \lim_{x \rightarrow0}  \frac{2a b\sin \frac x2\cos\frac x2\sqrt{a+b + \sqrt{a^2+b^2+ 2 ab \cos x}}}{\sqrt{ 2ab\cdot 2\sin^2\frac x2}} \\
=&  \lim_{x \rightarrow0} \sqrt{a b}\cos\frac x2 \sqrt{a+b + \sqrt{a^2+b^2+ 2 ab \cos x}} \\
=& \sqrt{a b}\sqrt{a+b + \sqrt{a^2+b^2+ 2 ab}} \\
=& \sqrt{2ab(a+b)} \\
\end{align}$$
Plug $L_1$ into (1)
$$\begin{align}
L=\frac{1}{2(a+b)}\cdot \sqrt{2ab(a+b)} = \sqrt{\frac{ab}{2(a+b)}} \\
\end{align}$$
