difference of square roots approximation

The approximation that you gave is in fact often more accurate. Multiply $\sqrt{x+a}-\sqrt{x+b}$ by $\dfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}$. We get $$\sqrt{x+a}-\sqrt{x+b}=\frac{a-b}{\sqrt{x+a}+\sqrt{x+b}}.\tag{$1$}$$ This is not an approximation, it is exact.

Moreover, when $x$ is very large in comparison to $|a|$ and $|b|$, the right-hand side of $(1)$ is computationally much better than the left-hand side. To see this, imagine we are doing the calculation on a calculator, or in floating point on a computer. If $x$ is very large, like $10^{10}$, adding $a$ to it on a calculator does not change the value, because of the limited precision. This means that our result may have a large relative error.

If $x$ is very large in comparison with $|a|$ or $|b|$, there is little relative error made if we replace $\sqrt{x+a}+\sqrt{x+b}$ in $(1)$ by $2\sqrt{x}$.

Factor out $x$ to get $$\sqrt{x}\left(\sqrt{1 + \frac{a}{x}} - \sqrt{1 + \frac{b}{x}}\right)$$ Now expand each square root in a Binomial series $$=\sqrt{x}\left(1 + \frac{a}{2x} - 1 - \frac{b}{2x} + \mathcal{O}(x^{-2})\right)$$ For large $x$ we can neglect the terms of order $x^{-2}$ and smaller to make the approximation $$\sqrt{x + a} - \sqrt{x + b} \approx \frac{a-b}{2\sqrt{x}}$$

Let $f(x)=\sqrt{x}$. Then, for $x>0$, we have $f(x+h) \approx f(x) + f'(x)h$. Hence $f(x+a)-f(x+b) = f(x+a)-f(x) - (f(x+b)-f(x)) \approx f'(x)(a-b)$. Since $f'(x) = \frac{1}{2\sqrt{x}}$, we have the desired approximation: $\sqrt{x+a}-\sqrt{x+b} \approx \frac{1}{2\sqrt{x}}(a-b)$.

The approximation comes from the Taylor theorem, and can be extended to higher orders.