Expanding √x at x = 4

1. With $u=\frac h4$ and $\sqrt{1+u}=1+\frac u2-\frac{u^2}8+\frac{u^3}{16}+o(u^3)$:
$$\sqrt{4+h}=2+\frac h4-\frac{h^2}{64}+\frac{h^3}{512}+o(h^3).$$

2. For $h=0.1$:
$$2+0.025-0.00015625+0.00000195=2.02484570.$$
The exact value is $2.02484567\dots$: the error is about $3\cdot10^{-8}$.

Remark. Always expand around the nearest "nice" point: the closer to the centre, the better the approximation. Expanding $\sqrt x$ at $0$ is impossible (the derivative blows up there).