Taylor expansion of $\sqrt{n-k}$

Question

I am reading a paper which casually assumes the asymptotic $\sqrt{n-k} \simeq \sqrt{n}-\frac{k}{2\sqrt{n}}$. This expression is what Wolfram calls Taylor expansion at infinity and from what I understand we work with $\sqrt{x-k}=\sqrt{x}\cdot \sqrt{1-k/x}$ and then we proceed on doing the Taylor expansion. My problem is that we also cannot do Taylor expansion near zero because said function is not differentiable there.
Another point is: do the Taylor expansion near point $1$ for $\sqrt{x}$ so that $$\sqrt{x}\sqrt{1-k/x} = \sqrt{x} - \frac{k}{2\sqrt{x}} - \frac{k^2}{8x^{3/2}}\frac{1}{\xi^{3/2}},$$ where $\xi$ as in Legendre residual form (between $1-k/x$ and $1$). We can then deduce that $\sqrt{n-k}\leq \sqrt{n}-k/2\sqrt{n},$ but what about the lower bound? Is there a way to fix an inequality relating the remainder and the second term of the above expansion?
My real question, is how to use the aforementioned fact in order to get a result of the form $$\sqrt{x-1/2}\geq \sqrt{x}\left(1-\frac{4}{x}\right).$$ The first two terms of $\sqrt{x-1/2}$ in the above expansion should be $\sqrt{x}-\frac{1}{4\sqrt{x}},$ so the remainder term should be $\geq -\frac{15}{4\sqrt{x}}$ or something.

Answer 1

If you want to show that $\sqrt{x-1/2}\geq \sqrt{x}\left(1-\frac{4}{x}\right),$

the following argument is simpler than bounding Taylor series terms:

(a) for $\frac12\le x<4$: $\sqrt{x-1/2}\geq0\gt \sqrt{x}\left(1-\frac{4}{x}\right);$

(b) for $x>\frac{32}{15}, \frac{15}2>\frac{16}x,$ so $x-\frac12>x-8+\frac{16}x=x(1-\frac4x)^2, $ so $\sqrt{x-1/2}>\sqrt x(1-\frac4x);$

so we have shown $\sqrt{x-1/2}>\sqrt x\left(1-\frac4x\right)$ for all $x$ where $\sqrt{x-1/2}$ is defined.