Why doesn't $\zeta(s) = \frac{1}{1-2^{1-s}}\eta(s)$ approach $\frac{-1}{2}$ as $s$ approaches $0$?

Question

According to Wikipedia (https://en.wikipedia.org/wiki/Riemann_zeta_function), an equation which extends the Riemann zeta function from $Re(s)>1$ to $Re(s)>0$ is an extension of the Riemann zeta function using the Dirichlet eta function

However, when I plot the function using only real arguments on Desmos, the function seemingly approaches $0$ as $x$ approaches $0$, which seems to contradict the statement that $\zeta(0) = \frac{-1}{2}$, which is in the same article.

Answer 1

Close to zero $$\zeta(s)=-\frac{1}{2}-\frac{1}{2} \log (2 \pi )\, s+O\left(s^2\right)$$ Your formula, extended to infinity, gives $\large\color{red}{-}\zeta(s)$

Answer 2

The behavior is merely an artifact of the truncation of the infinite series. Indeed, let

$$ \zeta_N(s) = \frac{1}{1-2^{1-s}} \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^s}. $$

Below demonstrates the graph of $\zeta_N(s)$ and $\eta(s)$ for $N=1,4,9,16,\ldots,10000$:

Indeed, the graph of $\zeta_N(s)$ does approach $\zeta(s)$ in the regime $s > 0$.

On top of this, $\lim_{N\to\infty} \zeta_N(s)$ converges only when $\operatorname{Re}(s)>0$, and in particular, it diverges when $s = 0$. Hence, the alternating behavior of $\zeta_N(0) $ cannot be used for any claim against the identity $\zeta(0) = -\frac{1}{2}$.