It is frequently stated that the continuum limit of a lattice model is equivalent to the low-energy limit, e.g. here, here and section IIB of this. I do not know how to show this for myself. Take for example the tight-binding Hamiltonian
$$ H = -t \sum_n c_{n+1}^\dagger c_n + \text{h.c.},\quad t \in \mathbb{R}$$
with periodic boundary conditions, where $\{ c_n \}$ are fermionic modes that obey $\{ c_n,c_m^\dagger \} = \delta_{nm}$, indices label lattice sites and h.c. denotes hermitian conjugate.
Continuum limit
We can express the lattice fermions as a continuous field sampled at discrete lattice points as $c_n \equiv \sqrt{a}c(x_n)$, where $x_n = na$ with lattice spacing $a$. We approximate the Hamiltonian to first order in $a$ as
$$ H = -t \sum_n a\left[ c^\dagger(x_n) + a\partial_x c^\dagger(x_n) + O(a^2)\right]c(x_n) + \text{h.c.} $$
Now take the continuum limit by setting $a \rightarrow 0$ to map the sums to integrals. We get
$$ H \approx -t \int dx c^\dagger(x) c(x) - v_F\int dx \left[ (\partial_x c^\dagger(x)) c(x)+c^\dagger(x) \partial_x c(x) \right] = -t \int dx c^\dagger(x) c(x) $$
where $v_F = 2at$ and going to the last equality I noted the second term can be written as a total derivative so integrates to zero.
Low energy limit
We Fourier transform the fields with the definition $ c_n = \frac{1}{\sqrt{N}} \sum_p e^{ipan} c_p$, in which case the Hamiltonian is diagonalised:
$$ H = \sum_p E(p) c^\dagger_p c_p, \quad E(p) = -2t\cos(ap)$$
The low-energy theory is described by expanding about the points where $E(p) = 0$. There are two unique points located at $p_\pm = \pm \frac{\pi}{2a}$ in the Brillouin zone. The low energy limit is therefore obtained as:
$$ H \approx \sum_k E(p_+ + k) c^\dagger_+(k) c_+(k) + E(p_- + k) c^\dagger_-(k)c_-(k) $$
where $c_\pm(k) = c_{p_\pm + k}$. Using $E(p_\pm + k) = \pm v_F k + O(a^2)$, we can inverse Fourier transform to find the Hamiltonian back in position space as
$$ H \approx H_+ + H_- ,\quad H_\pm = \pm v_F (-i) \int dx c^\dagger_\pm(x) \partial_x c_\pm(x)$$
This method is the method taken for deriving the continuum limit of graphene for example.
My question
It is clear that the continuum limit obtained by setting $a \rightarrow 0$ in position space and the low-energy limit by expanding about Fermi points $p_\pm$ in momentum space yield different results. Why is it commonly stated that the continuum limit is the same as the low-energy limit and why didn't it work for me?
