I calculated the polynomial or order $2$ for $\ln(x)$, centered at $x_o=1$, which is:
$$\ln(1.3) = \ln(1.0) + \ln'(1.0)(x-1) + \ln''(1.0)(x-1)^2$$
Where the lagrangian error is:
$$E(x) = \frac{\ln^3(x_a)}{3!}(1.3 -1)^3$$
where the special $x_a$ is in $[1.0, 1.3]$
I need to bound this error, so I must calculate its absolute value and then find it is less than a given quantity:
$$|E(x)| = \frac{3}{6*10^3}\frac{2}{|x_a^3|}$$
The maximum value that the error can take in the interval $[1.0, 1.3]$ is on $1.0$, so I can consider $x_a = 1.0$ as the worst error, therefore concluding that:
$$|E(x)| < \frac{1}{10^3}$$
but my book says $|E(x)|<10^{-2}$
There is an error in the second formula for $|E(x)|$. It should be $$|E(x)| =\frac{1}{6} \Bigl(\frac{3}{10}\Bigr)^3\frac{2}{|x_a^3|}\le\frac{9}{1000}<10^{-2}.$$