It's trivial to see that my linear approximation is continuous:
Let's start by examining the limit at \( \log_b^{\circ 2}(e) \) from the right:
\(
\begin{eqnarray}
{\Large \lim_{\normalsize z \to \log_b^{\circ 2}(e)^{+}}}\ \left[ \mathrm{slog}_b(z) \right]
& = & n + \frac{\log_b^{\circ 2}(e)-\exp_b^{\circ n}(1)}{\log_b(e)-\log_b^{\circ 2}(e)}
\end{eqnarray}
\)
Then from the left:
\(
\begin{eqnarray}
{\Large \lim_{\normalsize z \to \log_b^{\circ 2}(e)^{-}}}\ \left[ \mathrm{slog}_b(z) \right]
& = & \left. \mathrm{slog}_b\left(b^z\right) - 1\right|_{z=\log_b^{\circ 2}(e)} \\
& = & \mathrm{slog}_b\left(\log_b(e)\right) - 1 \\
\\[4pt]
\\
& = & n + \frac{\log_b(e)-\exp_b^{\circ n}(1)}{\log_b(e)-\log_b^{\circ 2}(e)} - \frac{\log_b(e)-\log_b^{\circ 2}(e)}{\log_b(e)-\log_b^{\circ 2}(e)}\\
\\[4pt]
\\
& = & n + \frac{\log_b^{\circ 2}(e)-\exp_b^{\circ n}(1)}{\log_b(e)-\log_b^{\circ 2}(e)} \\
\end{eqnarray}
\)
The limits at the other end of the interval are similarly simple to demonstrate.
Let's start by examining the limit at \( \log_b^{\circ 2}(e) \) from the right:
\(
\begin{eqnarray}
{\Large \lim_{\normalsize z \to \log_b^{\circ 2}(e)^{+}}}\ \left[ \mathrm{slog}_b(z) \right]
& = & n + \frac{\log_b^{\circ 2}(e)-\exp_b^{\circ n}(1)}{\log_b(e)-\log_b^{\circ 2}(e)}
\end{eqnarray}
\)
Then from the left:
\(
\begin{eqnarray}
{\Large \lim_{\normalsize z \to \log_b^{\circ 2}(e)^{-}}}\ \left[ \mathrm{slog}_b(z) \right]
& = & \left. \mathrm{slog}_b\left(b^z\right) - 1\right|_{z=\log_b^{\circ 2}(e)} \\
& = & \mathrm{slog}_b\left(\log_b(e)\right) - 1 \\
\\[4pt]
\\
& = & n + \frac{\log_b(e)-\exp_b^{\circ n}(1)}{\log_b(e)-\log_b^{\circ 2}(e)} - \frac{\log_b(e)-\log_b^{\circ 2}(e)}{\log_b(e)-\log_b^{\circ 2}(e)}\\
\\[4pt]
\\
& = & n + \frac{\log_b^{\circ 2}(e)-\exp_b^{\circ n}(1)}{\log_b(e)-\log_b^{\circ 2}(e)} \\
\end{eqnarray}
\)
The limits at the other end of the interval are similarly simple to demonstrate.
~ Jay Daniel Fox
