$\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$, $\theta$ must be in radians. But $x$ can be in degree for $\lim_{x\to\pm\infty}\frac{\sin x}{x}=0$? 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!
 A: 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.
A: 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.   
A: 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:
$$\lim_{y\rightarrow \infty}\frac{\operatorname{sind}(y)}{y}$$
