On the limit of $S(x)=\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}$ when $x\to 0^+$ Sorry I haven't any ideas.
Maybe it equals ${1 \over 2}$.
$$\lim_{x\to0^+}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}$$

As shown in the comment, the easiest way to evaluate the sum is using
Riemann zeta function:
$$\lim_{x\to 0^{+}}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x} = \lim_{x\to 0^{+}}(1-2^{1-x})\zeta(x) = (1-2)\zeta(0) = \frac12$$
However, it is possible to evaluate the sum without it. We can start from following integral representation of $\frac{1}{n^x}$,
$$\frac{1}{n^x} = \frac{1}{\Gamma(x)} \int_0^\infty t^{x-1} e^{-nt} dt$$
where $\Gamma(x)$ is the Gamma function
and derive an integral representation of the sum.
$$\begin{align}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x} 
&= \frac{1}{\Gamma(x)}\sum_{n=1}^\infty (-1)^{n-1}\int_0^\infty t^{x-1} e^{-nt} dt\\
&= \frac{1}{\Gamma(x)}\int_0^\infty \left( \sum_{n=1}^\infty (-1)^{n-1} e^{-nt}\right) t^{x-1} dt\\
&= \frac{1}{\Gamma(1+x)}\int_0^\infty \frac{1}{e^t + 1} d t^{x}\\
&= \frac{1}{\Gamma(1+x)}\int_0^\infty \frac{1}{e^{t^{1/x}} + 1} dt
\end{align}
$$
As long as $x > 0$ and we keep manipulating the terms in the sum in unit of pairs,
everything is non-negative and we won't have any issue of exchanging the order of summation
and integration.
In the limit of $x \to 0^{+}$, the integrand converges pointwisely to a step function $\theta(t)$:
$$\lim_{x\to 0^{+}} \frac{1}{e^{t^{1/x}} + 1} = \theta(t) \stackrel{\text{def}}{=}
\begin{cases} 
\frac12,& t \in [0,1)\\
\frac{1}{e+1}, & t = 1\\
0, &t \in (1,\infty)
\end{cases}$$
It is clear for small enough $x$, the integrand is dominated by some Lebesgue integrable
function over $\mathbb{R}$. We can then use
dominated converge theorem
to justify the exchange of order of taking limit and integration. As a result, we find
$$\lim_{x\to 0^{+}} \sum_{n-1}^\infty \frac{(-1)^{n-1}}{n^x} = \frac{1}{\Gamma(1)} \int_0^\infty \theta(t) dt = \int_0^1 \frac12 dt = \frac12$$
