Using Weierstrass’s Factorization Theorem I am trying to factorize $\sin(x)\over x$ which by Taylor series expansion and using the roots is $$a \cdot \left(1 - \frac{x}{\pi} \right) \left(1 + \frac{x}{\pi} \right) \left(1 -  \frac{x}{2\pi} \right) \left(1 + \frac{x}{2\pi} \right) \left(1 - \frac{x}{3\pi} \right) \left(1 + \frac{x}{3\pi} \right) \cdots$$
Now I was told that this nasty factor $a$ conveniently becomes $1$ due to Weierstrass’s Factorization Theorem which is a transcendental generalization of the Fundamental Theorem of Algebra.
My question
Could you please show me how $a$ is being neutralized using this theorem? Or don't you even need this theorem to do so?
 A: The value of this product for small x's is the product of $(1-x^2/(n \pi)^2)$ which, when you take logs (and due to the second power in x), behaves like the sum over n of $-x^2/(n\pi)^2$, which approaches 0 as x approaches 0.
A: The Weierstrass factorization theorem as usually stated tells you only that $a=e^{g(x)}$ for some entire function $g(x)$.  Hadamard's refinement says a little more, based on the growth rate of the function.  In your case, since $\left| \frac{\sin x}{x} \right| < \exp\left(|x|^{1+o(1)} \right)$ as the complex number $x$ grows, Hadamard tells you that $g(x)$ is a polynomial of degree at most $1$.  Since $\frac{\sin x}{x}$ and $\prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2} \right)$ are both even functions, so is $e^{g(x)}$.  Thus $g(-x)-g(x)$ is a (constant) integer multiple of $2\pi i$.  Hence $g(x)$ is constant, and so is $a$.  Finally, as everyone else has pointed out, taking the limit as $x$ goes to $0$ shows that $a=1$.
See Ahlfors, Complex analysis for more about Hadamard's refinement, which relates the "order" and "genus" of an entire function.
