TITLE: Limit involving Zeta and Gamma function
QUESTION [8 upvotes]: Can someone help me evaluate this limit?
$$\lim_{x\to +\infty}\frac {\zeta(1+\frac 1x)}{\Gamma(x)}$$
I never came across this kind of limit so I don't even know where to start.

REPLY [2 votes]: Good old L'Hôpital tells us $$\lim_{x\to\infty}\frac{\zeta\left(1+\frac{1}{x}\right)}{\Gamma(x)}=\lim_{x\to\infty}\frac{\zeta^{'}\left(1+\frac{1}{x}\right)}{\Gamma^{'}(x)}=\lim_{x\to\infty}\frac{\zeta^{'}\left(1+\frac{1}{x}\right)}{\Gamma(x)\psi^{(0)}(x)},$$ where $\psi^{(n)}$is the $n$-th derivative of the digamma function. 
Combining $\lim\limits_{x\to1}\zeta(x)=\lim\limits_{x\to+\infty}\Gamma(x)=\lim\limits_{x\to+\infty}\psi^{(0)}(x)=+\infty$ and $\lim\limits_{x\to1}\zeta'(x)=-\infty$ we deduce that for large enough $x$ the leftmost fraction is positive whereas the rightmost is negative, so if the limit exists it must be equal to $0$.

Alternatively, the result is straightforward rewriting $\zeta$ and $\Gamma$ by means of the Riemann functional equation, Stirling's approximation and the infinite product definition for the gamma function (due to Euler himself): $$\displaystyle\Gamma(x)=\frac{1}{x}\prod\limits_{n=1}^\infty\frac{\left(1+\frac{1}{n}\right)^x}{1+\frac{x}{n}}.$$