A new proof of $\left ( \lim_{x\rightarrow 0} {sin(x) \over x} \right )$ by using Euler reflection formula.

Euler reflection formula

$${\pi \over {sin(\pi a)} } = \Gamma (a)\Gamma (1-a)$$

where  $a\notin \mathbb{Z}$ 

let  $x = \pi a \Rightarrow a = {x \over \pi}$

So the formula can be written as:

$$\sin(x) = {\pi \over {\Gamma({x \over \pi}) \Gamma(1 - {x \over \pi})}}$$

where  ${x \over \pi}\notin \mathbb{Z}$

Thus, we have.

$$\lim_{x \rightarrow 0}{\sin(x) \over x} = \lim_{x \rightarrow 0}{1 \over {{x \over \pi}\Gamma({x \over \pi}) \Gamma(1 - {x \over \pi})}}$$

$$= \lim_{x \rightarrow 0}{1 \over {\Gamma({{x \over \pi} + 1}) \Gamma(1 - {x \over \pi})}}$$

$$= {1 \over {\Gamma({0 + 1}) \Gamma(1 - 0)}}  = {1 \over {\Gamma({1})^2}} = 1$$