<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] Since $8^2 = 64$ and $9^2 = 81$ and $81 - 64 = 17$, using $8.9$ as an estimate for $\sqrt {80}$ is more than just a lucky guess. Since 
$\quad (8.9)^2 < 80 \text{ and } (9)^2 > 80$
we must have
$\quad 0 \lt \sqrt{80} - 8.9 \lt 0.1$
Define the function
$\tag 1 F(x) = \frac{8.9x + 80}{x+8.9}$
For $x \ne -8.9$ we have the following identity,
$\tag 2 F(x) -  \sqrt{80} = (x - \sqrt{80}) \, (x + 8.9)^{-1} \, (8.9 - \sqrt{80}) $
(c.f. this link)
Since $(8.9 + 8.9)^{-1} \lt 0.1$ the following must be true
$$\tag 3 |F(8.9) -  \sqrt{80}| \lt (0.1)^3 = 0.001$$
All that remains is to evaluate $F(8.9)$,
$\tag 4 F(8.9) = \frac{15921}{1780}$
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