Reiteration: use radians for trig functions in mathematics to avoid an endless stream of fiddly constants to memorize.
Length of a circular arc:\begin{align*}
\text{radians: }& & L &= r \theta \\
\text{degrees: }& & L &= \frac{2 \pi r \theta}{360^\circ}
\end{align*}
Area of a sector of a circle:\begin{align*}
\text{radians: }& & A &= \frac{1}{2}\pi r^2 \theta \\
\text{degrees: }& & A &= \frac{\pi r^2 \theta}{360^\circ}
\end{align*}
(And the "$1/2$" in the radians version is easier to remember when you learn calculus because the integral of $r\theta$ with respect to $r$ is $\displaystyle \frac{1}{2}r^2\theta$. The "$1/2$"s in other places arise the same way: $\displaystyle \frac{1}{2}mv^2$, for instance.)
However, the limit also works for radians, but you have to change everything to radians. We change $\theta$, in radians, to $\phi$, in degrees, via $\theta = \frac{\pi}{180^\circ} \phi$ (introducing another constant to remember).
\begin{align*}
\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta}
&= \lim_{\frac{\pi}{{180\;}^\circ} \phi \rightarrow 0} \frac{\sin \frac{\pi}{180^\circ} \phi}{\frac{\pi}{180^\circ} \phi} \\
&= \lim_{\phi \rightarrow 0} \frac{\sin \frac{\pi}{180^\circ} \phi}{\frac{\pi}{180^\circ} \phi} \\
&\overset{\text{L'H}}{=} \lim_{\phi \rightarrow 0} \frac{\cos \left(\frac{\pi}{180^\circ}\phi \right) \cdot \frac{\mathrm{d}}{\mathrm{d}\phi} \frac{\pi}{180^\circ}\phi }{\frac{\pi}{180^\circ} } \\
&= \lim_{\phi \rightarrow 0} \frac{\cos(0) \cdot \frac{\pi}{180^\circ} }{\frac{\pi}{180^\circ} } \cdot \frac{\frac{180^\circ}{\pi}}{\frac{180^\circ}{\pi}} \\
&= \lim_{\phi \rightarrow 0} \frac{1 \cdot 1 }{1} \\
&= 1 \text{.}
\end{align*}
(Full Disclosure: I've used L'Hôpital's rule above. For this particular limit, to avoid circular reasoning, the derivative of the sine function is found using the radian version of this limit. If you prove only the degree version using the squeeze theorem, then the derivative of sine from this, the use of L'Hôpital's rule above is circular.)
