Lateral limits of function involving hyperbolic trignometric functions I am not being able to calculate the lateral limits at 0 of the following function
$f(x) = \frac{\sinh(x)}{2\sqrt{\cosh(x) - 1}}$
I have tried both direct substitution (yields 0/0) and L'Hospital's rule (yields (0/0)/0 )

Since
$\cosh(2x)
=2\sinh^2(2x)+1
$
and
$\sinh(2x)
=2\sinh(x)\cosh(x)
$,
$\begin{align*}
f(x) 
&= \frac{\sinh(x)}{2\sqrt{\cosh(x) - 1}}\\
&= \frac{\sinh(x)}{2\sqrt{2\sinh^2(x/2)}}\\
&= \frac{\sinh(x)}{2\sqrt{2}\sinh(x/2)}\\
&= \frac{2\sinh(x/2)\cosh(x/2)}{2\sqrt{2}\sinh(x/2)}\\
&= \frac{\cosh(x/2)}{\sqrt{2}}\\
&\to \frac{1}{\sqrt{2}}
\qquad\text{since } \cosh(x) \to 1 \text{ as }x \to 0\\ 
\end{align*}
$
