Please bear with me. So, for example the limit for $a$ in this hyperbola:
$$\lim_{a \to 0^+} (y^2 -x^2 = a) \, ,\,\,\, \, y \ge 0 \tag{A}$$
This appears to make the equation go to: $$y^2 = x^2 \, ,\,\,\, \, y \ge 0 \tag{B}$$ $$y = |x| \tag{C}$$
So, is $(A)$ a valid mathematical definition of the absolute value? And furthermore, is $(A)$ differentiable? Because if it was, that immediately means that it's differentiable as $a$ is aribitrarily close to $0$ which is the same thing as $(C)$, but we know $(C)$ is not differentiable at $0$.
In short, I'm wondering if (and maybe the above is a bad example) we can show, via a parameter, that one function converges to another in this type of limiting process and, if so, why would it lose it's differentiability (if that even makes sense)?
What you've stumbled upon is a property of something called uniform convergence. Uniform convergence is a way to talk about how a sequence of functions approaches a limiting function.
A sequence of functions $f_n(x) \xrightarrow[unif]{} f(x) $ if for each $\epsilon > 0$ and $x \in I$ there exists N such that $n > N$ implies $| f_n(x) - f(x) | < \epsilon$
What this means intuitively is that our sequence of functions converges to the limiting function at a worst case rate. So for all points in the interval, the error, $| f_n(x) - f(x) |$ can be made as small as we want.
Lets prove that your sequence converges uniformly: Let $f_n(x) = \sqrt{x^2 + \frac{1}{n}}$ and let $f(x) = |x|$. If we let $n \to \infty$, then this is identical to your question, except we only take the upper branch of the hyperbola.
Let $\epsilon > 0$ be given. Let $x \in I$, where $I$ is some bounded interval. Now, choose $N = \frac{1}{\epsilon^2}$ Then $n>N$ implies $$\left|\sqrt{x^2+\frac{1}{n}}-|x|\right|=\sqrt{x^2+\frac{1}{n}}-|x|$$ Since $x^2$ and $\frac{1}{n}$ are both positive$$\sqrt{x^2+\frac{1}{n}}-|x|\leq \sqrt{x^2}+\sqrt{\frac{1}{n}}-|x|$$ But $\sqrt{x^2}=|x|$ so we have $$\sqrt{\frac{1}{n}} \leq \sqrt{\frac{1}{N}}=\epsilon$$ So the sequence converges uniformly for ANY bounded interval.
What does all this mean? It means that you are exactly right. The upper branch of that hyperbola does approach $|x|$. Uniform convergence, then, must not conserve differentiability. It does however conserve continuity and integrability.