Example

We have $(1 + x)^{1/x} = \exp \left( \frac{1}{x} \log (1 + x) \right)$: when $x$ tends to $+\infty$ we have $\log (1 + x) = \log x + \log \left( 1 + \frac{1}{x} \right)$, whence the asymptotic expansion of $\frac{1}{x} \log (1 + x)$ with respect to the scale of the $x^\alpha (\log x)^\beta$:

From this expansion, and from the Taylor expansion

on a neighbourhood of $u = 0$, one deduces the asymptotic expansion

with respect to the scale $x^\alpha (\log x)^\beta$ by the methods explained above.

Keeping the same hypotheses and notation, the asymptotic expansion of $e^f$ does not pose new problems except when $f \gg 1$; one must then distinguish two cases, as $g_\alpha \gg 1$ or $g_\alpha \ll 1$. In the first case, giving an expansion for $f$ does not allow one to obtain a principal part for $e^f$ relative to $\mathscr{E}$, because one does not know in general whether the remainder $r_\alpha$ tends to 0, that is, whether $e^{r_\alpha}$ tends to 1. On the contrary, if $g_\alpha \ll 1$ then $r_\sigma \ll 1$ and so $e^f \sim \exp \left( \sum_{\lambda \leq \alpha} a_\lambda g_\lambda \right)$. One can make this result more precise: let $a_\gamma g_\gamma$ be the principal part of $f$ and let $\delta$ be the index (such that $\gamma < \delta \leq \alpha$) for which $g_\delta = 1$; we put $f_1 = \sum_{\lambda \leq \delta} a_\lambda g_\lambda$, $f_2 = \sum_{\delta < \lambda \leq \alpha} a_\lambda g_\lambda + r_\alpha$; we have $f = f_1 + f_2$, so $e^f = e^{f_1} e^{f_2}$, and the general method explained at the start of this subsection allows one to form an asymptotic expansion of $e^{f_2}$ (starting from the Taylor expansion of $e^t$ at the point $t = 0$). One will then have an asymptotic expansion for $e^f$ if $e^{f_1} = \prod_{\lambda \leq \delta} \exp(a_\lambda g_\lambda)$ belongs to $\mathscr{E}$, or to a scale $\mathscr{E}_1$ containing $\mathscr{E}$.