Limit of $\frac{\sin(\theta)}{\theta}$ in degrees

Question

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 $0.01745$.

Answer 1

$$ \lim_{\theta\to0}\frac{\sin\theta^\circ}{\theta^\circ} = \lim_{\theta\to0} \frac{\sin\left( \dfrac{\pi\theta}{180}\text{ radians} \right)}{\theta} = \lim_{\eta\to0} \frac{\sin(\eta\text{ radians})}{\left( \dfrac{180\eta}{\pi} \right)} = \frac\pi{180}\lim_{\eta\to0} \frac{\sin\eta}{\eta}. $$

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 $1$.

Answer 2

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

$$\lim_{x\rightarrow 0} \frac{2\pi}{360} \frac{\cos(\frac{2\pi x}{360})}{1} = \frac{2\pi}{360} \sim 0.01745.$$

Answer 3

If you know the limit in radians $$\lim_{x \to 0} \frac{\sin x}{x}$$ then the limit, with $x$ given in degrees is $$\lim_{x \to 0} \frac{\sin \frac{\pi x}{180}}{x} = \frac{\pi}{180}\lim_{x \to 0} \frac{\sin \frac{\pi x}{180}}{\frac{\pi x}{180}}$$

Answer 4

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".

Answer 5

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?"