$\omega(k)$ and $k(\omega)$ about waves Wave propagation is characterized by the wavenumber $k$ and the angular frequency $\omega$. Sometimes (like in this answer) the relation $\omega (k)$ is preferred; sometimes instead $k ( \omega)$ is used.
When expanding $k(\omega)$ around a point $k = k_0$:
$$\omega (k) \simeq \omega(k_0) + (k - k_0) \left. \frac{d\omega(k)}{d k} \right|_{k = k_0} + \frac{1}{2} (k - k_0)^2 \left. \frac{d^2 \omega(k)}{dk^2} \right|_{k = k_0}$$
or vice-versa:
$$k (\omega) \simeq k(\omega_0) + (\omega - \omega_0) \left. \frac{dk(\omega)}{d \omega} \right|_{\omega = \omega_0} + \frac{1}{2} (\omega - \omega_0)^2 \left. \frac{d^2 k(\omega)}{d\omega^2} \right|_{\omega = \omega_0}$$
$\left. \frac{d\omega(k)}{d k} \right|_{k = k_0} = v_g$ is the group velocity in the first expansion; instead $\left. \frac{dk (\omega)}{d \omega} \right|_{\omega = \omega_0} = 1/v_g$ is its reciprocal. 
But what does happen if $v_g \to 0$? Will the second expansion have an infinite first-order coefficient? The same question is for the second derivative quantities. The two expansions should be interchangeable, anyway they seem to be different.
 A: Imagine a plot $\omega(k)$ where, for some $k_0$, $v_g(k_0) = \left. \frac{\mathrm d\omega}{\mathrm dk} \right|_{k_0} = 0$.
Then the graph will have a horizontal tangent at $k=k_0$, like in this example:

The graph of $k(\omega)$ is just the mirror image, obviously it will have a vertical tangent at the respective point:

So yes, the first derivative diverges and the function $k(\omega)$ can not be expanded around this point.
A: Let us write the dispersion relation in a mathematically clearer version : 
$$ \omega = f(k)$$  where we note $f$ the dispersion function.
Say you want to compute the group velocity at $k_0$. It is defined as:
$$ v_g(k_0) \equiv \frac{df}{dk}(k_0) = \left.\frac{d\omega(k)}{dk}\right|_{k = k_0} \text{ (in your notation)}$$
Up to this point this is a definition and if $v_g(k_0) \rightarrow0$ it means that the derivative of the dispersion function $f$ at $k_0$ is $0$, which is completely possible.
Now let us look at the inverse dispersion function $g$ defined by:
$$k=g(\omega)$$
If $f$ is invertible on a small interval around $k_0$ (say $[k_0-\epsilon,k_0+\epsilon]$ with $\epsilon \in \mathbb{R}$ ), then in this interval we can write $g=f^{-1}$ and there is a unique $\omega_0$ in the interval that verifies $k_0 = g(\omega_0)$ and $\omega_0 = f(k_0)$. In the case that $\frac{df}{dk}(k_0) \neq 0$, you have the following formula for the derivative of $g$ :
$$\left.\frac{dk(\omega)}{d\omega}\right|_{\omega = \omega_0} =  \frac{dg}{d\omega}(\omega_0) = \frac{1}{\frac{df}{dk}(g(\omega_0))} = \frac{1}{\frac{df}{dk}(k_0)} = \frac{1}{v_g(k_0)}$$
Note that this formula is only valid when $v_g(k_0) = \frac{df}{dk}(k_0) \neq 0$. This means that when you say that the two terms should be interchangeable, this is not mathematically true. It is only the case if the dispersion function $f$ is invertible around $k_0$ and its derivative at $k_0$ non zero. This solves mathematically the paradox that you are evoking.
To sum it up, it means that you have to chose a definition for the group velocity, and the most common one to my knowledge is $ v_g \equiv\left.\frac{d\omega(k)}{dk}\right|_{k = k_0}$. If it goes to zero at a certain value $k_0$ it means that the dispersion function has a horizontal tangent at this point, and that the reciprocal dispersion function has an infinite slope at a given value $\omega_0$. This doesn't imply anything on definition of the group velocity, since it is a definition !
I hope my answer was clear enough.
Cheers!
