## 82. Math Puzzle 3/72

If $x$ is not too large in magnitude, the approximation formula holds:

$$
\sqrt{1+x} \approx 1+\frac{x}{2}
$$

The error $\Delta$ of this approximation formula is sought. Estimate how large $x$ may be at most so that the error remains less than 0.01. (since $\Delta$ is only small, $\Delta^{2} \approx 0$).
Given the error $\sqrt{1+x}=1+\frac{x}{2}+\Delta$, squaring it yields

$$
1+x=1+x+\frac{x^{2}}{4}+\left(2+\frac{1}{2}\right) \Delta+\Delta^{2} \quad\left(\Delta^{2}=0\right)
$$

thus

$$
\Delta=\frac{-x^{2}}{8\left(1+\frac{x}{2}\right)}=-\frac{x^{2}}{8}
$$

since $|x|$ is only small. For $\Delta=0.01$, it must be that $|x| \leq 0.28$. The negative sign of $\Delta$ indicates that the approximation formula always gives values that are too large.