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We know that $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ but $\theta$ must be in radians. My first question is what happen when $\theta$ is not in radian? Is it only because in the proof we use radian so $\theta$ must be in radians?

Then we also know that $\lim_{x\to\pm\infty}\frac{\sin x}{x}=0$. We can also prove it using sandwich theorem: $0\le|\frac{\sin x}{x}|\le|\frac{1}{x}|$, but here we are not using the fact that $x$ must be in radian. My next question is does that mean that $\lim_{x\to\pm\infty}\frac{\sin x}{x}=0$ also works even when $x$ is in degree?

If we see its pretty strange, when $x$ approaches $0$, there will be a limit only when $x$ is in radian, but when $x$ approaches neg/pos infinity, the limit exists regardless whether $x$ is in radian or degree.

Many thanks for the helps!

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  • $\begingroup$ I find it a bit strange, from the point of view of meta-mathematics, to consider $\sin x/x$ where $x$ is in degrees. Because, intuitively, $\sin x$ is length, so the units of $\sin x/x$ will be length/degrees, very strange to me. However, if $x$ is in radians, then $x$ is also length (the length of the arc) and so $\sin x/x$ is ratio between to length, and this does make sense to me... $\endgroup$ – Lior B-S Jul 30 '14 at 6:49
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Yes, when change from degree to radian. You multiply the angle by a extra constant (what is it?), which does not change the fact that the angle approaches $\pm\infty$.

Similarly, the limit $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$, when $x$ is given in degree, still exists. I left it for you to figure out what it is.

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The results are not strange because as x tends to zero, a*x also tends to zero given 'a' is finite constant.Similarly when x tends to infinity, again sin(x) function for any 'x'(whether radian or degree) is limited between -1 and 1 and dividing it(a finite value) with a x (that is very large or tending to infinite) will always tend to zero.

The above answer is good but I thought to give a general explanation.

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Generally the first thing you want to do when figuring out questions like this is to give distinct names to distinct ideas. So let's call $\operatorname{sinr}(x)$ the sine function when $x$ is in radians and $\operatorname{sind}(x)$ the sine function when $x$ is in degrees.

We know: $$\lim_{x\rightarrow 0} \frac{\operatorname{sinr}(x)}{x} = 1\tag{A}$$ by assumption and we know $$\operatorname{sinr}\left(r\cdot2\pi\right) = \operatorname{sind}\left(r \cdot 360\right) \tag{B}$$ by treating $r$ in revolutions. We want to find: $$\lim_{y\rightarrow 0}\frac{\operatorname{sind}(y)}{y} \tag{C}$$ First, apply (B) to (C) to change the degrees function to the radian function. To do this you have to set $y = r\cdot 360$ $$= \lim_{360r \rightarrow 0}\frac{\operatorname{sind}(360r)}{360r}$$ $$ = \lim_{360r \rightarrow 0}\frac{\operatorname{sinr}(2\pi r)}{360r}\tag{D}$$

Now we want to use (A) to solve (D), so we set $2\pi r = x$:

$$ = \lim_{360\frac{x}{2\pi} \rightarrow 0}\frac{\operatorname{sinr}(x)}{360\frac{x}{2\pi}}$$

and take the constant factor out of the limit: $$ = \frac{2\pi}{360}\lim_{x\frac{360}{2\pi} \rightarrow 0}\frac{\operatorname{sinr}(x)}{x}$$

and $\frac{360}{2\pi}x \rightarrow 0$ only when $x \rightarrow 0$, so: $$ = \frac{2\pi}{360}\lim_{x \rightarrow 0}\frac{\operatorname{sinr}(x)}{x}$$

And now you can apply (A) and finish it. The same approach can be used to figure out what is: $$\lim_{y\rightarrow \infty}\frac{\operatorname{sind}(y)}{y}$$

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