Limit of ratio between arc and chord in 0 Recently I've computed the ratio between arc and corresponding chord in a circle of given radius $r$ and noticed that it doesn't tend to $1$ as the angle $x$ approaches $0$ as expected, but to $\pi/180$.
I used the formulas of arc ($\pi×r×x/180$) and corresponding chord ($2×r×\sin(x/2)$) in terms of angle and radius, where the radius is fixed and the angle is variable.
I used the fundamental limit $\lim_{x\to 0}x/\sin(x)=1$ and found that $$\lim_{x\to 0}\frac{\text {arc}}{\text{chord}}=\lim_{x\to 0}\frac{\pi×r×x/180}{2r×\sin(x/2)}=\lim_{x\to 0}(
\frac{\pi/180×(x/2)}{\sin(x/2)},$$ which tends to $\pi/180$.
Now, maybe it theoretically does make sense, but the good sense doesn't tell me the same thing. I mean, this whole thing I proved means that by making the angle (or the arc) close enough to $0$, we reach a point where the arc becomes smaller than the chord, which doesn't seem natural. Can someone explain me the link between the common sense and theoretical result, or whether I have a fault in interpretation or anything else? Thanks in advance!
Edit: $x$ is in degrees.
 A: You must distinguish the sine function that assumes its argument in degrees rather than in radians, and 
$$\lim_{d\to0}\frac{\sin_° d}d=\lim_{d\to0}\frac{\sin \dfrac{\pi d}{180}}d=\frac\pi{180}.$$
[$\sin_°$ denotes the sine applied to an angle in degrees.]
A: I presume you are using $x$ in degrees when you say that arc length is $r \cdot x \cdot \frac{\pi}{180}$. Lets be clear and rather write it as $x^\circ$.
The problem lies on the line where you say $
\lim_{x\to 0}\frac{\sin(x^\circ)}{x} = 1$. This is not true. The limit $\lim_{x\to 0}\frac{\sin(x)}{x} = 1$ if only angle $x$ is in radians.
Otherwise you may convert it into radians by multiplying with $\frac{\pi}{180}$:
$$\lim_{x\to 0}\frac{\sin(x^\circ)}{x} = \lim_{x\to 0}\frac{\sin(x \cdot \frac{\pi}{180})}{x} = \frac{\pi}{180}$$
A: Common sense and theoretical result are in agreement. 
Working in radians presents no problem$$\lim_{x\to 0}\frac{arc}{chord}=\lim_{x\to 0}\frac{\pi\times r \times \frac{x}{\pi}}{2r\times \sin(\frac{x}{2})}=\lim_{x\to 0}\frac{r\times x}{2r\times \sin(\frac{x}{2})}=\lim_{x\to 0}\frac{\frac{x}{2}}{\sin(\frac{x}{2})}=\frac{1}{1}$$
When working in degrees, however, recall that $\pi=3.14159...$, the length of the semi-circumference of a circle measured in radii (radians), is equal to $180$, the same length measured in degrees. Hence$$3.14159...\neq180$$ but $$3.14159...radians = 180 degrees$$(Similarly, $1\neq1000$ but $1$ kilometer = $1000$ meters.)
We can work in radians or degrees but not both at once. Working in degrees we get finally$$\lim_{x\to 0}\frac{arc}{chord}=\frac{\pi}{180}=\frac{180}{180}=\frac{1}{1}$$ Arc and chord approach equality as $x$ tends to $0$.
