In this answer, why is the $x$ in denominator not converted into degrees? This has been answered in the comments saying that $x$ is just a number and it only makes a difference in the $\sin(x)$ argument. But I am not satisfied with this answer. It doesn't convince me. Is there a more compelling answer to this question?
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Did you read the final comment, which interprets $x$ as arc length? – rogerl Sep 14 '20 at 14:27
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$$\frac{\sin\left(60^{o}\right)}{60}\ne\frac{\sin\left(\frac{\pi}{3}\right)}{\left(\frac{\pi}{3}\right)}$$
Hope this leaves no confusion left
Soumyadwip Chanda
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To see things from another point of view, we have that as $\theta_d\to 0$ (in degrees) numerically
$$\frac{\sin(\theta_d)}{\theta_d}\to 1 $$
but this result is not dimensionless therefore we need to multiply the solution by $\frac{\pi}{180}$ in order to obtain a dimensionless value.
user
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It's Just because the question is about the limit of $\frac{\sin \theta}{\theta}$ when $\theta$ is in degrees. However the well-known limit is with $\theta$ in degrees, so it is converted in radians for the sine.
More precisely, denote $\sin$ the sine function with the argument in radians and $\sin_\circ$ the sine when the argument is in degrees. We want the limit of $\;\dfrac{\sin_\circ\theta}{\theta}$ and we know the limit of $\dfrac{\sin\theta}\theta$($\theta$ in radians).
We very naturally convert $\sin_\circ$ in terms of $\sin$, i.e. to calculate the sine of $\theta$ (in degrees) we have to convert $\theta$ in radians first: $$\sin_\circ\theta=\sin\frac{\pi\theta}{180},$$ but in the given fraction, the denominator remains in degrees. Why should we have to convert it?
Bernard
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