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. 
 A: 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.
A: 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 $\sqrt{\frac{ab}{2(a+b)}}$.
A: 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*}
A: 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)
$$\begin{align}
L=\frac{1}{2(a+b)}\cdot \sqrt{2ab(a+b)} = \sqrt{\frac{ab}{2(a+b)}} \\
\end{align}$$
A: 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 $$\frac{ab} {2(a+b)}\sqrt {\frac{2(a+b)}{ab}}=\sqrt{\frac{ab}{2(a+b)}}$$
