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 )
 A: Write $\sinh x$ as $\frac{e^x-e^{-x}}2$ and $\cosh x$ as $\frac{e^x+e^{-x}}2$, then substitute $t=e^x$.
$$\begin{align}\lim_{x\to0}\frac{\sinh x}{2\sqrt{\cosh x-1}}&=\lim_{x\to0}\frac{\frac{e^{x}-e^{-x}}2}{2\sqrt{\frac{e^x+e^{-x}}2-1}}\\&=\lim_{t\to1}\frac{\frac{t^2-1}{2t}}{2\sqrt{\frac{t^2-2t+1}{2t}}}\\&=\lim_{t\to1}\frac{(t+1)(t-1)\sqrt{2t}}{(t-1)4t}\\&=\lim_{t\to1}\frac{(t+1)\sqrt{2t}}{4t}\\&=\frac{\sqrt2}{2}\end{align}$$
EDIT
I forget to put absolute value brackets to $t-1$ after canceling square root, so two-sided limit doesn't exist. For $x\to0^-$ limit will be $-\frac{\sqrt{2}}2$ and for $x\to0^+$ limit will be $\frac{\sqrt{2}}2$.
A: I can suggest for you this way of finding this limit, using the Taylor series expansion of  $\sinh(x)$  and $\cosh(x)$ near zero, to the order one . we have  $$\lim_{x\rightarrow 0} \frac{\sinh(x)}{2\sqrt{\cosh(x)-1}} = \lim_{x\rightarrow 0} \frac{x}{2\sqrt{1+ x^2/2-1}} =\frac{1}{\sqrt{2}} $$
A: 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*}
$
