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I found a proof for $\left( \lim\limits_{x\rightarrow 0} {\sin(x) \over x} \right)$ using Euler reflection formula

I would like to ask if this proof can be strong mathematical proof?

so let me start by 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:

$$\sin(x) = {\pi \over {\Gamma\left({x \over \pi}\right) \Gamma\left(1 - {x \over \pi}\right)}}$$ where ${x \over \pi}\notin \mathbb{Z}$

Thus, we have.

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

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

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

Kroki
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Pentapolis
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    I don't know if it is legit (seems correct reasoning though) but this is surely an approach I have never seen before. – Zima May 14 '23 at 21:27
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    Seems totaly legit, when defining $\Gamma$, proving this formula or even proving the residue theorem you don't need to know that $\sin x \sim x$ at $0$. So this is not circular ! Quite a weird proof aha – Zag May 14 '23 at 21:42
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    I might be wrong, but in order to show the reflection formula, don't you need to first prove the infinite product for $\frac{sinx}{x}$, which at some point uses the fact that the limit of $\frac{sinx}{x}$ goes to $1$? Again, I'm not sure if this is true – Fotis May 14 '23 at 21:47
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    Or if not the infinite product of $\frac{sinx}{x}$, then the power series expansion of $sinx$, which uses the derivative of $sinx$, hence the limit of $\frac{sinx}{x}$ as $x$ goes to $0$ – Fotis May 14 '23 at 22:00
  • @Fotis this is not the only proof of the reflection formula. There's one that uses the residue theorem. – Zag May 14 '23 at 22:38

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