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. 

Assume $u=\sqrt{a^+b^2+2ab\cos x} $ so that $u\to a+b$ as $x\to 0$. The expression under limit is $$\frac{ab\sin x} {2u\sqrt{a+b-u}}$$ whose limit is same as that of $$\frac{ab} {2(a+b)}\frac{x}{\sqrt{a+b-u}}$$ Next we can observe that $$\frac{a+b-u} {x^2}=\frac{(a+b) ^2-u^2}{x^2(a+b+u)}=\frac{2ab}{a+b+u}\cdot\frac{1-\cos x}{x^2}\to\frac{ab}{2(a+b)} $$ It follows that the desired limit is $$\frac{ab} {2(a+b)}\sqrt {\frac{2(a+b)}{ab}}=\sqrt{\frac{ab}{2(a+b)}}$$
