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. 

Fix $a,b > 0$, and let
$$
f(x)
=
\frac{ab\,\sin x}{2\sqrt{(a^2+b^2+2ab\cos x)(a+b-\sqrt{a^2+b^2+2ab\cos x})}}
$$
As $x$ approaches zero from the right, we have $f(x) > 0$, so the limit from the right, if it exists, $L$ say, must be nonnegative, in which case, since $f$ is an odd function, the limit from the left will be equal to $-L$.

Looking ahead, we'll find that $L > 0$, hence the two-sided limit doesn't exist.

Let's find $L$ . . .

Computing $L^2$, we get
\begin{align*}
L^2&=\lim_{x\to 0} f(x)^2
\\[4pt]
&=\lim_{x\to 0}\, \frac{(ab\,\sin x)^2}{4(a^2+b^2+2ab\cos x)(a+b-\sqrt{a^2+b^2+2ab\cos x})}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{\sin^2 x}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{\left(x\left(\!{\Large{\frac{\sin x}{x}}}\!\right)\right)^2}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{x^2}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{x^2}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\cdot \frac{a+b+\sqrt{a^2+b^2+2ab\cos x}}{a+b+\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{4(a+b)^2}\lim_{x\to 0}\, \frac{x^q}{a+b-\sqrt{a^2+b^2+2ab\cos x}}
\cdot \frac{2(a+b)}{a+b+\sqrt{a^2+b^2+2ab\cos x}}
\\[4pt]
&=\frac{(ab)^2}{2(a+b)}\lim_{x\to 0}\, \frac{x^2}{2ab-2ab\cos x}
\\[4pt]
&=\frac{ab}{4(a+b)}\lim_{x\to 0}\, \frac{x^2}{1-\cos x}
\\[4pt]
&=\frac{ab}{4(a+b)}\lim_{x\to 0}\, \frac{2x}{\sin x}
\\[4pt]
&=\frac{ab}{4(a+b)}\lim_{x\to 0}\, 2
\\[4pt]
&=\frac{ab}{2(a+b)}
\\[8pt]
\text{hence}\;L&=\sqrt{\frac{ab}{2(a+b)}}
\\[4pt]
\end{align*}
