I have to find a neighborhood of $x=0$ such that $\sin{x}$ is approximated with an error less than $10^{-n}$.
I have thought that from lagrange error I have:$$|\sin{x}-x|\leq \frac{|x|^3}{6}<10^{-n}$$So the neighbourhood is $[-(\frac{6}{10^n})^{\frac{1}{3}}, (\frac{6}{10^n})^{\frac{1}{3}} ]$Am I right?
