Limit of $\frac{\sin(\theta)}{\theta}$ in degrees What does $\lim \limits_{\theta\to0}\dfrac{\sin(\theta)}{\theta}$ equal when $\theta$ is expressed in degrees?
I know that theta in degrees is $\frac{\pi}{180}$ theta radians, but I don't get the final answer of $0.01745$.
 A: $$
\lim_{\theta\to0}\frac{\sin\theta^\circ}{\theta^\circ} = \lim_{\theta\to0} \frac{\sin\left( \dfrac{\pi\theta}{180}\text{ radians} \right)}{\theta} = \lim_{\eta\to0} \frac{\sin(\eta\text{ radians})}{\left( \dfrac{180\eta}{\pi} \right)} = \frac\pi{180}\lim_{\eta\to0} \frac{\sin\eta}{\eta}.
$$
This is why radians are used: When radians are used then $\lim\limits_{\eta\to0} \dfrac{\sin \eta} \eta=1$.
Postscript in response to comments below:
The question is how to find
\begin{align}
& \lim_{\theta\to0} \frac{\text{sine of $\theta$ degrees}}\theta = \lim_{\theta\to0}\frac{\text{sine function in radians}\left(\dfrac{\pi\theta}{180}\right)} \theta \\[12pt]
= {} & \frac\pi{180}\lim_{\theta\to0} \frac{\text{sine function in radians}\left(\dfrac{\pi\theta}{180}\right)} {\dfrac{\pi\theta}{180}} = \frac\pi{180}\lim_{\eta\to0} \frac{\text{sine function in radians}(\eta)}\eta.
\end{align}
In other words the notation "$\sin$" means a particular function: the one that maps a number $\eta$ to the sine of $\eta$ radians.
End of postscript
It's the same as the reason why $e$ is the "natural" base for exponential functions:
\begin{align}
\frac{d}{dx} 2^x & = (2^x\cdot\text{constant}) \\[10pt]
\frac{d}{dx} 3^x & = (3^x\cdot\text{a different consant}) \\[10pt]
\frac{d}{dx} 20^x & = (20^x\cdot\text{yet another constant})
\end{align}
etc.  Only when the base is $e$ is the "constant" equal to $1$.
A: The questions seems to be asking you to evaluate
$$\lim_{x\rightarrow 0} \frac{\sin^*(x)}{x},$$
where $\sin^*$ means the usual $\sin$ function but evaluating its argument in terms of degrees.
Now, since $360^\circ = 2\pi$ radians, we have the identity
$$\sin^*(x) = \sin\left(\frac{2\pi x}{360}\right)$$
(This is worth contemplating for a bit)
Anyway, we now are trying to evaluate
$$\lim_{x\rightarrow 0} \frac{\sin(\frac{2\pi x}{360})}{x}$$
Applying L'Hopital, this is equal to the limit
$$\lim_{x\rightarrow 0} \frac{2\pi}{360} \frac{\cos(\frac{2\pi x}{360})}{1} = \frac{2\pi}{360} \sim 0.01745.$$
A: If you know the limit in radians $$\lim_{x \to 0} \frac{\sin x}{x}$$ then the limit, with $x$ given in degrees is
$$\lim_{x \to 0} \frac{\sin \frac{\pi x}{180}}{x} = \frac{\pi}{180}\lim_{x \to 0} \frac{\sin \frac{\pi x}{180}}{\frac{\pi x}{180}}$$
A: When angles are said to be in degrees, this actually means that when applying trigonometric functions, they are interpreted as degrees. Hence in the question "$\sin(\theta)$" must be understood as $$\sin\left(\frac{\pi\theta}{180}\right)$$ where $\sin$ is the usual sine function (arguments in radians) and you are in fact computing
$$\lim_{\theta\to0}\frac{\sin\left(\dfrac{\pi\theta}{180}\right)}\theta.$$
There is no reason to convert at the denominator, division is still the ordinary division. Only the trigonometric function is "special".
A: The question is not properly posed.
In whatever units the is angle chosen the result is the same $(=1).$
For all units (degree, radian)
$$ lim_{X \rightarrow 0}\dfrac{\sin X}{X}$$
equals unity.
You should get that answer independently  when asked " how many radians are there in a degree?"
