Limit of sinx/x without taylor series or areas I learned at school that the fact that sinx/x goes 1 when x approaches 0 can be proved by comparing areas.
I'm wondering if there is another way to prove it. Do you have any idea about it?
 A: In the diagram, $x$ is the signed length of the arc $DB$, and $\sin x$ is the signed length of the segment $AB$ and $\tan x$ is the signed length of the segment $DC$.

Either $0<\sin x<x<\tan x$ or $\tan x<x<\sin x<0$.
In either case, $0<\dfrac{\sin x}{x}<1<\dfrac{\tan x}{x}=\dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}$.
Thus $\cos x<\dfrac{\sin x}{x}<1$.
Since $\lim_{x\to0}\cos x=1$ we have $\lim_{x\to0}\dfrac{\sin x}{x}=1$ by the "squeezing" theorem.
ADDENDUM Why is $x<\tan x$ when $0<x<\frac{\pi}{2}$?
Consider the following diagram:

Construct the tangent to the unit circle at $B$ and mark its intersection with the tangent line $CD$ as $E$. Then $E$ is equidistant from both $D$ and $B$. Construct the circle with center $E$ and radius $BE$. Then we have $x=\text{ arc }DB<BE+ED<DC=\tan x$ See link
Note that $C$ lies on a tangent line to the smaller circle and is different from the point of tangency, thus it lies outside the smaller circle.
A: There can be another simple intuitive proof for understanding purpose only. Consider a triangle with one angle x at O. And hypotenuse 1. Then the perpendicular gives the magnitude of sin x. Now just imagine the angle x is very small. Then what, the perpendicular would be vary small and the base is nearly the length of hypotenuse. Now to approximate you can consider perpendicular to be a small arc of circle, with radius as the base(=which is same as hypotenuse of unit length), centre at O. Now, using  the definition of angle, we can write: x= (arc length/radius)= sin x. Hence, the limit sin x/x tends to 1 for very small x. Hope this helps you in some way!
