In graphene, we have (in the low energy limit) the linear energy-momentum dispersion relation: $E=\hbar v_{\rm{F}}|k|$. This expression arises from a tight-binding model, in fact $E =\frac{3\hbar ta}{2}|k|$ where $t$ is the nearest-neighbor hopping energy and $a$ the interatomic distance. But how does one know that the Fermi velocity is given by $v_{\rm{F}} = \frac{3ta}{2}$? Normally one would use $m_{\rm{eff}}^{-1} =\frac{1}{\hbar^{2}} \frac{\partial^{2}E}{\partial k^{2}}$ and $v_{\rm{F}} = \frac{\hbar k_{\rm{F}}}{m_{\rm{eff}}}$, but in this case $m_{\rm{eff}} = \infty$.
[Physics] Effective Mass and Fermi Velocity of Electrons in Graphene:
condensed-matterelectronic-band-theorygraphenesolid-state-physics
Related Solutions
In calculating the electron dispersion you probably obtained the diagonalized Hamiltonian in the momentum space
$$ H=\sum_\mathbf{k}\left[c^{\dagger}_{\mathbf{k}A},c^{\dagger}_{\mathbf{k}B}\right]\left[\begin{array}{cc}0 & \Delta(\mathbf{k})\\ \Delta^{\dagger}(\mathbf{k}) &0\end{array}\right]\left[\begin{array}{c}c_{\mathbf{k}A} \\ c_{\mathbf{k}B}\end{array}\right]. $$
If you you chose your $x$ axis along the zigzag direction (arXiv:1004.3396), the two nonequivalent Dirac valleys are $\mathbf{K}_\kappa=\left(\kappa\frac{4\pi}{3\sqrt{3}a},0\right)$, $\kappa=\pm1$ and $\mathbf{K}_{-1}=\mathbf{K}^{\prime}$, where $a$ is the C-C distance. Then $\Delta(\mathbf{k})=-t\left(1+e^{-i\mathbf{k}\cdot\mathbf{a}_1}+e^{-i\mathbf{k}\cdot\mathbf{a}_2}\right)$, where $t$ is the hopping term, and $\mathbf{a}_1=\left(\sqrt{3}a/2,3a/2\right)$ and $\mathbf{a}_2=\left(-\sqrt{3}a/2,3a/2\right)$ are the lattice vectors.
Taylor expanding $\Delta(\mathbf{k})$ up to linear terms around those two points you obtain
$$ \Delta(\mathbf{k})=\kappa\frac{3ta}{2}q_x-i\frac{3ta}{2}q_y $$
where $\mathbf{q}$ is the displacement momenta from the $\mathbf{K}_\kappa$ point. Promoting these displacement momenta to operators you obtain the Hamiltonian
$$ H=\hbar v_F\left[\begin{array}{cccc}0 & q_x-iq_y & 0 & 0\\q_x+iq_y & 0 & 0 & 0\\0 & 0 & 0 & -q_x-iq_y\\0 & 0 & -q_x+iq_y & 0\end{array}\right] $$
where $v_F=\frac{3ta}{2\hbar}$ is the Fermi velocity. This is in $\left[\Psi_{A\mathbf{K}},\Psi_{B\mathbf{K}},\Psi_{A\mathbf{K}^{\prime}},\Psi_{B\mathbf{K}^{\prime}}\right]^T$ basis, if you rearrange your basis as $\left[\Psi_{A\mathbf{K}},\Psi_{B\mathbf{K}},\Psi_{B\mathbf{K}^{\prime}},\Psi_{A\mathbf{K}^{\prime}}\right]^T$ you get the compact form
$$ H=\hbar v_F\tau_z\otimes\boldsymbol{\sigma}\cdot\mathbf{k} $$
where $\tau_z$ acts in the valley space. This is similar to the Dirac-Weyl equation for relativistic massless particles, where instead of $v_F$ you get the speed of light
$$ H=\pm\hbar c\boldsymbol{\sigma}\cdot\mathbf{k} $$
where $+$ denotes right-handed antineutrions, and $-$ denotes left-handed neutrions. The differences are that $\boldsymbol{\sigma}=\left(\sigma_x,\sigma_y\right)$ for graphene acts in pseudospin space and $\boldsymbol{\sigma}=\left(\sigma_x,\sigma_y,\sigma_z\right)$ for neutrinos acts in real spin space.
Best Answer
The group velocity $v_g$ of a wave packet (that's the speed of the maximum of the wave packet) is given by $v_g=\frac{\partial\omega}{\partial k}$. In this case, $\frac{\partial\omega}{\partial k}=\frac 1 \hbar\frac{\partial E}{\partial k}$, which easily evaluates to $v_g=\frac{3ta}{2}=:v_f$ for $k=0$. That's actually the definition of $v_f$: it is the group velocity at $k=K$ ($K$ is the point in the Graphene bandstructure where the Dirac cone occurs - note that it is a vector because $k$ has an $x$ and a $y$ component), because $E(K)=E_f$.
The effective mass from solid state physics is indeed infinite. If one talks about "zero effective mass Dirac fermions" in Graphene, this comes from the massless Dirac equation which has the same dispersion relation. The solid state physics effective mass doesn't work here, because the dispersion relation needs to be parabolic (not linear with a cusp), there are two papers about that on arXiv here and here.