Area Swept by particle under central forces is an approximation

From Kepler’s second law, we infer, the conservation of angular momentum is equivalent to saying the areal velocity is constant,

And the proof goes like this
$$ mr^2{dottheta=L}
$$
where $L$ is angular momentum.
$$
frac{d(frac{1}{2} r^2dottheta)}{dt}={0} qquad …(1) (as L and m are constant)
$$

from the figure we can write
$$ dA= frac{1}{2}r r dtheta qquad..(2)$$
and from (1) and (2) we get
$$
frac{mathrm{d}A}{mathrm{d}t}=frac{1}{2}r^2frac{mathrm{d}theta}{mathrm{d}t}
$$

now since angular momentum is constant, we say, The radius vector sweeps out equal areas in equal times.

My problem is,

(1). To calculate the area element, we are taken an approximation, so how we are getting accurate results, (means we have equated it to angular momentum only after approximation and not in general)?

(2) Is Kepler’s second Law true up to some approximation? Because at the end we have just cared about the approximated triangle, but there remains a smaller region (as depicted in the figure) that we have not included. And at the planetary level, our approximation will not work?

(3) Can I say that it is not an approximation and I have missed something?

End Note: The 3 questions are actually 1 question, just written in points for the convenience of reader. I know the policy.