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. 

You can do it brutally with a limited developement at order 3:
Let's study the parts independantly and start with the denominator:


*

*$a^2+b^2+2ab\cos(x) = (a+b)^2-{abx^2}+O(x^4)$

*From that you have $\sqrt{a^2+b^2+2ab\cos(x)} = a+b - \frac{x^2 a b}{2 (a + b)} 
+ O(x^4)$

*Hence $a+b-\sqrt{a^2+b^2+2ab\cos(x)} = \frac{x^2 a b}{2 (a + b)} 
+ O(x^4)$

*Then: $(a^2+b^2+2ab\cos(x))(a+b-\sqrt{a^2+b^2+2ab\cos(x)}) = \frac{(a+b)ab}{2}x^2-\frac{a^2b^2}{2(a+b)}x^3+O(x)^4$

*Taking the square root gives a denominator in $x\sqrt{2(a+b)ab}+ o(x^2)$
But the numerator is a $abx+O(x^3)$ so that we have:
$f(x) = \frac{ab}{\sqrt{2(a+b)ab}} +o(1) = \sqrt{\frac{ab}{2(a+b)}} +o(1)$
All in all we therefore have a limit at $0$, which is $\sqrt{\frac{ab}{2(a+b)}}$.
