Benedict Irwin edited Investigation.tex  over 9 years ago

Commit id: 19b54afde24ba6f2c1534184d6ff60fa75eb9be7

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\begin{equation}  \lim_{n\to\infty} \frac{1|3;n}{0|8;n} = 1.5\overline{0} =\frac{12}{8} \\  \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}  Which would at least imply \begin{equation}  \lim_{n\to\infty} \frac{1|3;n}{d|8;n}=\frac{12}{(d|8) - d}  \end{equation}