Given the following facts:
A Banach space $X$ is Hilbert if and only if, for evert $x,y\in X$ we have $$2\Vert x\Vert^2+2\Vert y\Vert^2=\Vert x+y\Vert^2+\Vert x-y\Vert^2$$
The category $\textbf{Ban}_1$ of Banach spaces and contractions is complete and cocomplete
Suppose that for every $\varepsilon>0$ we have a Banach space $H_\varepsilon$ that is an $\varepsilon$ short of being Hilbert in the sense that, for every $x,y\in H_\varepsilon$ we have $$\left|2\Vert x\Vert^2+2\Vert y\Vert^2-\Vert x+y\Vert^2-\Vert x-y\Vert^2\right|<\varepsilon$$
Is there some limit of the $H_\varepsilon$ (or a countable subset of them) in $\textbf{Ban}_1$ that is a Hilbert space $H$? If so, can we use it as a method of proof? If none/all of the $H_\varepsilon$ satisfy some property $\mathbf{P}$ then, maybe (depending on how the property and the limit behave) we could say that $H$ does/does not satisfy $\mathbf{P}$.
I suspect the idea might work better if we instead ask for $$\alpha_\varepsilon\left(\Vert x+y\Vert^2+\Vert x-y\Vert^2\right)\le 2\Vert x\Vert^2+2\Vert y\Vert^2\le \beta_\varepsilon\left(\Vert x+y\Vert^2+\Vert x-y\Vert^2\right)$$ $$\left|\frac{\alpha_\varepsilon}{\beta_\varepsilon}-1\right|<\varepsilon$$ and maybe also ask for a "valid" limit $H$ to come with bicontinuous bijective linear maps $T_\varepsilon:H_\varepsilon\to H$ such that, for every $x\in H_\varepsilon$ we have $$A_\varepsilon\Vert x\Vert_{H_\varepsilon}\le\Vert T_\varepsilon x\Vert_H\le B_\varepsilon\Vert x\Vert_{H_\varepsilon}$$ $$\left|\frac{A_\varepsilon}{B_\varepsilon}-1\right|<\varepsilon$$ so that $H$ and $H_\varepsilon$ are close in some sense.
Maybe the details need to be totally distinct but the base idea is to have a property for a "sequence" of Banach space and hope for that property to hold also for the "limit" of that sequence, which we ask to be a Hilbert space. Could this work?
Thanks