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. 

Let us denote by $\pi - x$ the angle between the sides $a$ and $b$ from a triangle with third side $c(x)$, the application of the cosine's law results into
\begin{align*}
c^{2}(x) = a^{2} + b^{2} - 2ab\cos(\pi - x)
\end{align*}
Thus the given expression can be rewrriten as
\begin{align*}
f(x) = \frac{ab\sin(x)}{2\sqrt{c^{2}(x)(a + b - \sqrt{c^{2}(x)})}} = \frac{ab\sin(x)}{2c(x)\sqrt{a+b-c(x)}}
\end{align*}
Moreover, the expression $ab\sin(x)/2$ is the area of such triangle.
According to the Heron's formula, we have that
\begin{align*}
A = \frac{1}{4}\sqrt{(a+b+c)(a + b - c)(a - b + c)(-a + b + c)}
\end{align*}
Consequently, one has that
\begin{align*}
f(x) & = \frac{ab\sin(x)\sqrt{(a+b+c(x))(a - b + c(x))(-a + b + c(x))}}{2c(x)\sqrt{(a+b+c(x))(a + b - c(x))(a - b + c(x))(-a + b + c(x))}}\\\\
& = \frac{\sqrt{(a+b+c(x))(a - b + c(x))(-a + b + c(x))}}{4c(x)}
\end{align*}
Therefore we have that
\begin{align*}
\lim_{x\rightarrow 0}f(x) & = \frac{\sqrt{(a+b+c(0))(a-b+c(0))(-a+b+c(0))}}{4c(0)} = \sqrt{\frac{ab}{2(a+b)}}
\end{align*}
and we are done.
