--- In

primenumbers@yahoogroups.com,

Sebastian Martin Ruiz <s_m_ruiz@...> wrote:

> Limit {x->0} (Zeta[2+(x!)^x]-Zeta[3])Log[x]/(x(Zeta[3]-Zeta[2+x^x]))=EulerGamma=0.57721...

Surgeon General's warning: quitting Mathematica

/now/ greatly reduces serious risks to your brain.

I assume that x! means gamma(1+x). In which case, the value

of the limit has /nothing/ to do with the Riemann zeta function.

Try replacing "zeta" by "exp", or "log", any other function

that is analytic in the neighbourhood of z=3, and you will

get the /same/ result.

Explanation: gamma(1+x) = 1 - Euler*x + O(x^2)

David