Limit involving Zeta function 
Evaluate
$$\displaystyle \lim_{x \to 1} (x-1) \zeta (x)$$

I'm not much familiar with the properties of zeta function. An elementary solution is appreciated.
Thanks.
 A: The detail of answer depends on how you define $\zeta(s)$, but if you are working with the definition
$$ \zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^s}, \quad s > 1 $$
then notice that the following inequality
$$ \frac{1}{s-1} = \int_{1}^{\infty} \frac{dx}{x^s} \leq \zeta(s) \leq 1 + \int_{1}^{\infty} \frac{dx}{x^s} = \frac{s}{s-1} $$
immediately yields
$$ \lim_{s \downarrow 1} (s-1)\zeta(s) = 1. $$
A: Let me expand on the hint given in this answer.

The zeta function is the sum of the series
$$
\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}
$$
The sum of the even terms is
$$
2^{-s}\zeta(s)=\sum_{n=1}^\infty\frac1{(2n)^s}
$$
For any series, subtracting twice the sum of the even terms from the total sum gives the alternating sum:
$$
\left(1-2^{1-s}\right)\zeta(s)=\sum_{n=1}^\infty(-1)^{n-1}\frac1{n^s}
$$
Therefore, applying L'Hôpital,
$$
\begin{align}
\lim_{s\to1^+}(s-1)\zeta(s)
&=\lim_{s\to1^+}\frac{s-1}{\left(1-2^{1-s}\right)}\lim_{s\to1^+}\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^s}\\
&=\frac1{\log(2)}\sum_{n=1}^\infty(-1)^{n-1}\frac1n\\[6pt]
&=1
\end{align}
$$
