If the rocket maintains the same rest mass and also the same thrust in the center of mass system, its terminal velocity can be as close to the speed of light as you wish. It will just take forever.
It is interesting to ask when a given velocity is reached. Say the thrust remains the same all the time in a stationary frame. The mass of the rocket will grow like $m_0 \gamma$ where $\gamma = 1/\sqrt{1 - v^2/c^2}$ is the Lorentz factor. Then you can write
$\dot p = F$ where $p$ is the momentum and $F$ is the thrust. Also $p = m v = \gamma(v) m_0 v$. This leads to the equation
$$F = \dot p = m_0 \frac{\mathrm d}{\mathrm d t} \gamma(v(t)) \, v(t) \,.$$
Inserting the values, one has the differential equation
$$\frac{\mathrm d}{\mathrm d t} \gamma(v(t)) \, v(t) = \frac{100 \, \mathrm N}{100 \, \mathrm{kg}} := a_0 $$
which one has to solve for $v(t)$ with $v(0) = 0$. As the other side does not depend on time explicitly, one can just integrate both sides from $t' = 0$ to $t' = t$. Mathematica gave the this solution:
$$ v(t) = \frac{a_0 c t}{\sqrt{a_0^2 + c^2 t^2}} \,. $$
This is the code that I have used:
gamma[v_] = 1/Sqrt[1 - v^2/c^2]
DSolve[{D[gamma[v[t]] v[t], t] == a0, v[0] == 0}, v, t]
This is the corresponding plot:
You can see that the acceleration is constant (linear part) at first. Then your rocket drive has diminishing returns and it will asymptotically approach the speed of light.