The Dirichlet Eta Function and the limits interchange 1-) Let us handle the Dirichlet Eta Function and its partial sumshere which is convergent in the region
$0<\sigma< 1$  with  $s= \sigma +it $
$$ \eta (s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}},  \text{  and...... }\eta_N (s)=\sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{s}} $$
Meanwhile, we have already known that:
$$\eta (s_0)=0 ⇔ \eta (1-s_0)=0$$
.
Can we interchange the following limits?  If yes, how we can prove it ?
$$ \tag{1}\lim_{s\to s_0}\lim_{N\to \infty}\sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{s}}=\lim_{N\to \infty}\lim_{s\to s_0}\sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{s}} $$

2-) Let us suppose we have the following equality in the region
$0<\sigma< 1$
$$\tag{2}\lim_{s\to s_0}\frac {\lim_{N\to \infty}\eta_N (s)}{\lim_{N\to \infty} \eta_N (1-s)}=h(s_0)$$
$h(s)$ is continuous and finite and non-zero in the region
$0<\sigma< 1$
Thus, can we write the following equality depending on equation (2)? If yes, how we can prove it ?
$$\tag{3}\lim_{N\to \infty}\frac {\eta_N (s_0)}{ \eta_N  (1-s_0)}=h(s_0)$$
 A: For $\Re(s) > 0$
$$|\eta(s)-\eta_{2N}(s)| = |\sum_{n=N}^\infty (2n+1)^{-s}-(2n+2)^{-s}|=|\sum_{n=N}^\infty \int_{2n+1}^{2n+2} s x^{-s-1}dx|\\\le |s|\sum_{n=N}^\infty \int_{2n}^{2n+2} x^{-\Re(s)-1}dx=|s|\frac{(2N)^{-\Re(s)}}{\Re(s)}$$
(this proves $\eta(s)$ is well-defined, continuous, analytic, bounded by $|s|/\Re(s)$)
If $\eta(s)\ne 0$ and $\Re(s)\in (0,1)$ then $$\lim_{N \to \infty} \frac{\eta_N(s)}{\eta_N(1-s)} = \frac{\eta(s)}{\eta(1-s)}$$
If $\eta(s)=0$ it is different : $$|\eta(s)-\eta_{2N}(s)-\frac12 (2N+1)^{-s}|\\ = |\sum_{n=N}^\infty \int_0^1 s((2n+x+1)^{-s-1}-(2n+2x+1)^{-s-1})dx|\\=
 |\sum_{n=N}^\infty \int_0^1 s\int_x^{2x} (s+1)(2n+t+1)^{-s-2}dtdx|\le \frac{|s(s+1)| (2N+1)^{-\Re(s)-1}}{\Re(s+1)}$$
thus for $\eta(s)=0,\Re(s) \in (1/2,1)$ $$\lim_{N \to \infty} \frac{\eta_N(s)}{\eta_N(1-s)} = \lim_{N \to \infty} \frac{\frac12(2N+1)^{-s}}{\frac12 (2N+1)^{s-1}}=0$$
A: @reuns, you wrote as: 
If $\eta(s)\ne 0$ and $\Re(s)\in (0,1)$ then $$\lim_{N \to \infty} \frac{\eta_N(s)}{\eta_N(1-s)} = \frac{\eta(s)}{\eta(1-s)}=h(s)$$
Thus now, let us consider some $s$ which very near $s_0$ as $ε$ sufficiently small as $0<ε$  as below: 
$$s_0^-=(σ_0-ε)+it_0 \text{  and} ......s_0^+=(σ_0+ε)+it_0$$
$$\tag{4}\lim_{s\to s_0^-}\frac {\lim_{N\to \infty}\eta_N (s)}{\lim_{N\to \infty} \eta_N (1-s)}=h(s_0^-)$$
$$\tag{5}\lim_{N\to \infty}\frac {\eta_N (s_0^-)}{ \eta_N  (1-s_0^-)}=h(s_0^-)$$
We could write it also with $s_0^+=(σ_0+ε)+it_0$
Anyway now, we can see easly the follwing one by  $\lim {ε\to 0}$.
$$h(s_0^-)=h(s_0^+)=h(s_0)$$
Because   $h(s)$ is continuous and finite and non-zero in the region 
$0<\sigma< 1$
Thus, can we write the following one depending equation (4) ? According to me YES !!.
$$\tag{6}\lim_{N\to \infty}\frac {\eta_N (s_0)}{ \eta_N  (1-s_0)}=h(s_0)$$
