Benedict Irwin edited Investigation.tex  over 9 years ago

Commit id: da41813c4568389152f5bb767f2bd5ddab2d281c

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\end{equation}  This doesn't sound, too ridiculous when one considers that $0.|9;\infty=1$ However one can notice that \begin{equation}  \lim_{n\to\infty} \frac{1|3;n|8}{2|8;n|7}=\lim_{n\to\infty} \frac{1|3;n}{2|8;n}  \end{equation}  In fact the end digits wil be very small, so as we go to infinity, this information will probably be lost.  \begin{equation}  \lim_{n\to\infty} \frac{1|3;n}{1|8;n} = 0.\overline{7058823529411764} =\frac{12}{17} \\  \lim_{n\to\infty} \frac{1|3;n}{2|8;n} = 0.\overline{461538} =\frac{12}{26} \\  \lim_{n\to\infty} \frac{1|3;n}{3|8;n} = 0.3\overline{428571} =\frac{12}{35} \\  \lim_{n\to\infty} \frac{1|3;n}{4|8;n} = 0.\overline{27} =\frac{12}{44} \\  \end{equation}