When you write the solution of the overdamped system as $$ \eqalign{ & f(t) = c_{\,1} e^{\,\rho \,t + \omega \,t} + c_{\,2} e^{\,\rho \,t - \omega \,t} = \left( {c_{\,1} e^{\,\omega \,t} + c_{\,2} e^{\, - \omega \,t} } \right)e^{\,\rho \,t} = \cr & = \left( {a\cosh \,\left( {\omega \,t} \right) + b\sinh \,\left( {\omega \,t} \right)} \right)e^{\,\rho \,t} \cr} $$
and impose the initial conditions, for instance for $f(0)$ and $f'(0)$, you get $$ \left\{ \matrix{ f(0) = a \hfill \cr f'(0) = \,\,\rho a + b\omega \quad \Rightarrow \quad b = \;{1 \over w}\left( {f'(0) - \,\,\rho f(0)} \right) \hfill \cr} \right. $$
so $$ f(t) = \left( {f(0)\cosh \,\left( {\omega \,t} \right) + {1 \over w}\left( {f'(0) - \,\,\rho f(0)} \right)\sinh \,\left( {\omega \,t} \right)} \right)e^{\,\rho \,t} $$
Now, if the damping approaches the critical value, that is $\omega \to 0$, then
$$ \bbox[lightyellow] { \mathop {\lim }\limits_{\omega \, \to \,0} f(t) = \left( {f(0) + \left( {f'(0) - \,\,\rho f(0)} \right)t} \right)e^{\,\rho \,t} }$$
And similarly, starting from an under-damped system, where you have normal $\sin $ and $\cos$ instead of the hyperbolic version.
So the critically-damped response is at the frontier between the two, mathematically and physically, and not easy distinguishable at first sight when very near to the critical value.
In fact, the under-damping case will always be evidenced by more or less visible oscillations.
In case of overdamping instead, since for passive system $\rho$ is negative, the exponential decay is prevalent and masking in the long time. In the short time instead, the difference between $\cosh (\omega t)$ and $1$ and $\sinh (\omega t)$ and $\omega t$ is not appreciable.
