Limit involving Zeta and Gamma function

Hint: Note that for $x>1$, we have $$\zeta \left(1+\frac{1}{x}\right) = \frac{1}{1^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{3^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\frac{1}{5^{1+\frac{1}{x}}}+\cdots< \\ \frac{1}{1^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\cdots = \\ 1+\frac{2}{2^{1+\frac{1}{x}}}+\frac{4}{4^{1+\frac{1}{x}}}+\frac{8}{8^{1+\frac{1}{x}}}+\cdots = \\ 1+\frac{1}{2^{\frac{1}{x}}}+\frac{1}{4^{\frac{1}{x}}}+\frac{1}{8^{\frac{1}{x}}}+\cdots= \\ (2^{-\frac{1}{x}})^0+(2^{-\frac{1}{x}})^1+(2^{-\frac{1}{x}})^2+(2^{-\frac{1}{x}})^3+\cdots = \\ \frac{1}{1-2^{-\frac{1}{x}}}$$

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 $0$.

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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): $$\displaystyle\Gamma(x)=\frac{1}{x}\prod\limits_{n=1}^\infty\frac{\left(1+\frac{1}{n}\right)^x}{1+\frac{x}{n}}.$$