Solving the Limit of sin(\pi*n/4)*\Gamma(x) Problem

[SumThePrimes]

I've recently been confronted with the limit as n goes to 0 of sin(\pi*n/4)*\Gamma(x) , and have no idea on how to confront the problem, as I have little familiarity with the gamma function. Is there any relatively easy ways to prove this, or at least ways that use methods not difficult to learn? I would very much like to see a proof, as wolfram alpha gives a answer of \pi/4, and the answer is important relating to some very interesting alternating series.
edit 2: Wow, I haven't solved this problem, but if wolfram alpha is right, soon I'll be summing alternating series never summed before :) Well, ones I've never seen summed before at least...

 

[morphism]

$\lim_{n \to 0} \sin(\frac{\pi n}{4}) \Gamma(x) = \sin(0)\Gamma(x) = 0$...

 

[mathman]

I suspect there is a typo in the original question - it should be Γ(n), not Γ(x).

 

[Citan Uzuki]

It's actually quite easy. You know that the gamma function satisfies the functional equation \Gamma(x+1) = x \Gamma(x), so then:

\begin{align*} \lim_{x \rightarrow 0} \sin (\pi x/4) \Gamma(x) &= \lim_{x \rightarrow 0} \frac{\sin(\pi x/4) \Gamma(x+1)}{x} \\ &= \left( lim_{x \rightarrow 0} \frac{\sin(\pi x/4)}{x} \right) \left( \lim_{x \rightarrow 0} \Gamma(x+1) \right) \\ &= lim_{x \rightarrow 0} \frac{\sin(\pi x/4)}{x}\end{align*}

Where the last equality follows from the fact that \Gamma is continuous and \Gamma(1) = 1. But then by L'hopital's rule:

\lim_{x \rightarrow 0} \frac{\sin(\pi x/4)}{x} = \lim_{x \rightarrow 0} \frac{\pi}{4} \cos (\pi x/4) = \frac{\pi}{4}