Proofs of $\lim_{x\to 0}\frac{\sin(x)}{x}=\lim_{x\to 0}\Gamma(x)\sin(x)=1$

Question

Well it's a little question:

How do you prove that:

$$\lim_{x\to 0}\frac{\sin(x)}{x}=\lim_{x\to 0}\Gamma(x)\sin(x)=1$$

Where we speak about the Gamma function.

The most quick proof uses the reflection Euler formula because:

$$\lim_{x\to 0}\Gamma(x)\sin(x)=\lim_{x\to 0}\Gamma(x)\Gamma(1-x)\sin(x)=1$$

Well I think there are a lot of proofs so I add the tag "big-list".

Particularly I would be happy to see a proof using the squeeze theorem.

Thanks in advance!!!

Answer 1

For latter part $$\lim_{x\to 0}\Gamma(x)\sin x=\lim_{x\to 0} \Gamma(x)\left[x\cdot \frac{\sin x}{x}\right]=\lim_{x\to 0} x\Gamma(x)=\lim_{x\to 0}\Gamma(x+1)=\Gamma(1)=1$$ By definition of functional equation of Gamma function $x\Gamma (x)=\Gamma(1+x)$.

Or by reflection formula

$$\lim_{x\to 0}\Gamma(x)\sin x =\lim_{x\to 0} \left[\frac{\pi}{\sin{\pi x}\cdot \Gamma(1-x)} \right]\cdot \sin x =\lim_{x\to 0}\left[\frac{\pi x}{x\sin(\pi x)\Gamma(1)}\right]\cdot \sin x= \lim_{x\to 0} \frac{\sin x}{x\Gamma(1)} =1$$

Answer 2

The Legendre Duplication Formula $$\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2^{1-2z} \; \sqrt{\pi} \; \Gamma(2z),\quad \text{Re } z > 0,$$

and

$$\Gamma(z + 1) = z\,\Gamma(z),\quad \text{Re } z > 0,$$

gives

$$\Gamma(x)\sin(x)=\frac{2^{1-2x} \; \sqrt{\pi} \; \Gamma(2x)\sin(x)}{\Gamma\left(x + \frac{1}{2}\right)}=\frac{2^{1-2x} \; \sqrt{\pi} \; \Gamma\left(2x+1\right)\sin(x)}{\Gamma\left(x + \frac{1}{2}\right)2x}.$$

Then, taking the limit as $x\to 0$ with the well-known $$\Gamma(n)=(n-1)!,\quad\Gamma(1/2)=\sqrt{\pi},$$ forms

$$\lim_{x\to 0}\Gamma(x)\sin(x)=\lim_{x\to 0}\frac{2\,\sqrt{\pi}~\Gamma(1)\sin(x)}{\sqrt{\pi}~2x}=\lim_{x\to 0}\frac{\sin(x)}{x}=1.$$