TITLE: Is it necessary to have $\theta$ in radians to obtain $\frac{\sin{\theta}}{\theta} \to 1$ as $\theta\to 0$?
QUESTION [7 upvotes]: $$\lim_{\theta \to 0} \frac{\sin{\theta}}{\theta} = 1$$
Could anyone explain why $\theta$ is in radians?
Is it because if $\theta$ is not in radians then $\frac{\sin{\theta}}{\theta} \neq 1$?
Is there a detailed explanation?

REPLY [4 votes]: $$
\lim_{x\to0}\frac{\sin x^\circ}{x} = \lim_{x\to0}\frac{\sin_\text{(radians)}\left(\frac{\pi x}{180}\right)}{x}.
$$
Now let $u=\dfrac{\pi x}{180}$.  Notice that as $x\to0$, we have $u\to0$.  And $x=\dfrac{180u}{\pi}$.  So the limit becomes
$$
\lim_{u\to0}\frac{\sin_\text{(radians)}\left( u \right)}{ \left( \frac{180u}{\pi} \right) } = \lim_{u\to0} \frac{\pi}{180} \cdot \frac{\sin_\text{(radians)} u}{u} = \frac{\pi}{180} \cdot \lim_{u\to0} \frac{\sin_\text{(radians)} u}{u} = \frac{\pi}{180}\cdot 1= \frac{\pi}{180}.
$$
Therefore
$$
\lim_{x\to0}\frac{\sin x^\circ}{x} = \frac{\pi}{180}.
$$
Only when radians are used is the limit $1$.
Likewise, only when radians are used is $\sin'$ equal to $\cos$.  If degrees are used then $\sin' = \frac{\pi}{180}\cdot\cos$.
This is similar to the situation with derivatives of exponential functions:
$$
\frac{d}{dx} a^x = \left(a^x\cdot\text{constant}\right).
$$
Only when $a=e$ is the constant equal to $1$.  That's what's "natural" about $e$.
Later note: I'd add a graphic to back this up if I could conveniently draw it and upload it.  But here I'll describe it.  First look at this: http://upload.wikimedia.org/wikipedia/commons/4/45/Unitcircledefs.svg Then imagine what happens if $\theta$ is infinitely small but positive.  In that case, the curved arc you're looking at looks like a straight line, and the vertical line that defines the sine is the same straight line.  So the ratio of their lengths is $1$.  But if you use some units besides radians, so that the length of the arc is not measured in the same units as the length of the vertical line, then their lengths are two different numbers, even though their lengths are the same, so the ratio of those numbers is not $1$.