We have  \( \lim_{\theta\to0} { \sin\theta \over \theta } \) = 1

Consider the below diagram.

We have

r = radius of the circle.
A = centre of the circle.
The sector ⌔ formed by the arc BD subtends an angle θ at the centre.

Case 1 : θ > 0 i.e. θ is +ve

Let 0 ≤ θ ≤ \( \pi \over 2\)

area △AOB = \( {1 \over 2} × BD × OA \) = \( {r \over 2} BD \)

area △AOB = \( {BD \over OB } \) = \( {\sin\theta} \)

⇒ BD =  \(r \, {\sin\theta} \)

∴ area △AOB = \( {1 \over 2} \,r × r  {\sin\theta} \) = \( {1 \over 2} r^2 \sin\theta \)

area ⌔ AOB = \( {\theta \over 2?}  ×    ? \, r^2 \)  = \(   { 1 \over 2 } r^2 \theta \)

area △AOC = \( {1 \over 2} × OA × CA \) = \(   { 1 \over 2 } r CA \)

also, \( { AC \over OA } = \tan \theta \)

⇒ AC = OA \( \tan\theta)\) = \( r tan\theta \)

∴ area △AOC = \( {1 \over 2} r^2 \tan\theta \)

Case 2 : θ < 0 i.e. θ is -ve

In this case, let θ = -\( \beta \)

then

\( \lim_{\theta\to0} { \sin\theta \over \theta} \) = \( \lim_{\beta\to0} {\sin (-\beta) \over -\beta} \)

= \( \lim_{\beta\to0} {\sin \beta \over \beta} \) = 1

Thus, 

\(\mathbf{lim_{θ\to0} } \mathbf{sinθ \over θ } \) = 1