<s>[INST] Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$ Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$
I thought it could be good to use the function $f: ]-\infty, 81] \rightarrow \mathbb{R}$ given by $f(x) = \sqrt{81-x}$
Because this function is of class $C^{\infty}$, we can compute its Taylor expansion given by :
$$T^n_{0} = 9 - \frac{1}{2}(81)^{-1/2}(x) + \frac{1}{4}(81)^{-3/2}\frac{x^2}{2}-\frac{3}{8}(81)^{-5/2}\frac{x^3}{6}\ + \dotsm $$
By the Lagrange remainder, $\exists$ for each $n \in \mathbb{N}$, $c_n \in [0,1]$ such that :
$$R^n(1) = f^{n+1}(c) \frac{1-0^{n+1}}{(n+1)!} = f^{n+1}(c) \frac{1}{(n+1)!} \leq 9.\frac{1}{(n+1)!} \lt 10^{-3}$$ 
$=> n(+1)! \gt \frac{9}{10^{-3}} = 9000$
So we can take $n = 8$
The approximation seems a little bit tricky to calculate especially without a calculator. I'm wondering if everything above is correct ? 
 [/INST] Barry Cipra's answer does everything that is required with no fuss at all, so it gets my vote, but I was still curious to know how much work it would be to get the result using the binomial series. Not a lot, I think:
$$
1 - \sqrt{1 - x} = \tfrac{1}{2}\sum_{k=0}^\infty a_kx^{k+1} \quad (|x| < 1),
\qquad a_k = \frac{1\cdot3\cdots(2k-1)}{4\cdot6\cdots(2k+2)},
\qquad a_0 = 1, \ a_1 = \tfrac{1}{4}.
$$
Because $a_{k+1} < a_k$, we can estimate the remainder using the
geometric series:
$$
0 < 1 - \sqrt{1 - x} - \tfrac{1}{2}\sum_{k=0}^{p-1} a_kx^{k+1} <
\frac{a_px^{p+1}}{2(1 - x)} \quad (p \geqslant 0, \ 0 < x < 1),
$$
In the present case, it is enough to take $p = 1$:
$$
0 < 1 - \frac{x}2 - \sqrt{1 - x} < \frac{x^2}{8(1 - x)} \quad (0 < x < 1).
$$
Taking $x = \frac1{81},$ and multiplying throughout by $9,$ we get $x/(1 - x) = \frac1{80},$ and so
$$
0 < \frac{161}{18} - \sqrt{80} < \frac1{8 \times 9 \times 80} = \frac1{5760}.
$$
In fact,
\begin{align*}
\frac{161}{18} - \sqrt{80} & \bumpeq 0.0001725, \\
\frac1{5760} & \bumpeq 0.0001736.
\end{align*}
</s>