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 $0.01745$.
[Math] Limit of $\frac{\sin(\theta)}{\theta}$ in degrees
calculuslimits
Best Answer
$$ \lim_{\theta\to0}\frac{\sin\theta^\circ}{\theta^\circ} = \lim_{\theta\to0} \frac{\sin\left( \dfrac{\pi\theta}{180}\text{ radians} \right)}{\theta} = \lim_{\eta\to0} \frac{\sin(\eta\text{ radians})}{\left( \dfrac{180\eta}{\pi} \right)} = \frac\pi{180}\lim_{\eta\to0} \frac{\sin\eta}{\eta}. $$
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 $1$.