How does $\sin\theta=\theta$ give a right answer even when it is an approximation?

Question

Magnetic field at the center of circular current carrying loop is given by

$$ B=\frac{\mu I}{2 R} $$

Where $\mu$ is the permeability of free space and $R$ is the radius of loop.

In a question by calculation I got a field at the center of the regular $n$-sided polygon as

$$ B=\frac{\mu I n \sin \frac \pi n} {2 \pi R} $$

Now the question also asks about what when $n$ goes to infinity the polygon is a circle.

With some assumptions

1. $n$ goes to infinity, hence $\theta$ goes to zero
2. hence $\sin \theta=\theta$

then we get the magnetic field same as a circle.

My question is when we derive $B$ directly assuming a circle we get an answer. when we derive $B$ of polygon and assume $n=\infty$ and do approximation, why do we get the same correct answer since there is an approximation involved.

the $\theta$ for a 4 sided polygon is shown

Answer 1

Instead of seeing your magnetic field by assuming $\sin\theta = \theta$ for small angles, you can rewrite your magnetic field as \begin{equation} B=\frac{\mu In\sin{\frac{\pi}{n}}}{2 \pi R} = \frac{\mu I}{2R} \frac{n\sin{\frac{\pi}{n}}}{\pi}=\frac{\mu I}{2R} \frac{\sin{\frac{\pi}{n}}}{\frac{\pi}{n}} \end{equation} In the limit $n \rightarrow \infty$ or $\frac{\pi}{n} = \theta \rightarrow 0$, $\lim_{\theta \rightarrow 0}$ $\frac{\sin{\theta}}{\theta}=1$ and hence $B=\frac{\mu I}{2R}$.

This approximation of taking $\theta \rightarrow 0$ as $n \rightarrow \infty$ is a valid one and gives an exact answer.

Answer 2

The source of your confusion is that $n=4$ is not very far along the path to your limit $n\to\infty$. For a twenty-sided polygon (shown) the difference between $\theta$ and $\sin\theta$ starts in the third significant figure.