We have  $\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }$ = 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 ≤ $\frac{\pi }{2}$

area △AOB = $\frac{1}{2}\times BD\times OA$ = $\frac{r}{2}BD$

area △AOB = $\frac{BD}{OB}$ = $\sin \theta$

⇒ BD =  $r\sin \theta$

∴ area △AOB = $\frac{1}{2}r\times r\sin \theta$ = $\frac{1}{2}{r}^2\sin \theta$

area ⌔ AOB = $\frac{\theta }{2𝛑}\times 𝛑r^2$  = $\frac{1}{2}{r}^2\theta$

area △AOC = $\frac{1}{2}\times OA\times CA$ = $\frac{1}{2}rCA$

also, $\frac{AC}{OA}=\tan \theta$

⇒ AC = OA $\tan \theta )$ = $rtan\theta$

∴ area △AOC = $\frac{1}{2}{r}^2\tan \theta$

Case 2 : θ < 0 i.e. θ is -ve

In this case, let θ = -$\beta$

then

$\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }$ = $\underset{\beta \to 0}{lim}\frac{\sin (-\beta )}{-\beta }$

= $\underset{\beta \to 0}{lim}\frac{\sin \beta }{\beta }$ = 1

Thus, 

${\mathbf{lim}}_{\mathbf{\theta }\to \mathbf{0}}\frac{\mathbf{sin}\mathbf{\theta }}{\mathbf{\theta }}$ = 1