Find the order of the infinitesimal alpha(x) = e

Find the order of the infinitesimal $\alpha(x) = e-\bigl(1+x\bigr)^{1/x}$ with respect to $x$ as $x\to0$. Recall that $\alpha(x)$ is of order $n$ with respect to $x$ if $\displaystyle\lim_{x\to0}\frac{\alpha(x)}{x^{n}} = C\neq0$.

Step 1: Pass to the logarithm

\[ \bigl(1+x\bigr)^{1/x} = e^{\,f(x)}, \qquad f(x) = \frac{\ln(1+x)}{x}. \]

Step 2: Expand $f$

From $\ln(1+x) = x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-\dfrac{x^{4}}{4}+O\!\left(x^{5}\right)$, dividing by $x$: \[f(x) = 1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+O\!\left(x^{4}\right) = 1+g(x), \tag{1}\]

\[ g(x) = -\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+O\!\left(x^{4}\right). \tag{2} \]

Step 3: Expand the exponential

Since $\bigl(1+x\bigr)^{1/x} = e\cdot e^{g(x)}$, expand $e^{g} = 1+g+\dfrac{g^{2}}{2}+\dfrac{g^{3}}{6}+O\!\left(g^{4}\right)$ with

\[ g^{2} = \frac{x^{2}}{4}-\frac{x^{3}}{3}+O\!\left(x^{4}\right), \qquad g^{3} = -\frac{x^{3}}{8}+O\!\left(x^{4}\right). \tag{3} \]

Collecting powers: the coefficient of $x$ is $-\dfrac12$; of $x^{2}$, $\dfrac13+\dfrac18 = \dfrac{11}{24}$; of $x^{3}$, $-\dfrac14-\dfrac16-\dfrac{1}{48} = -\dfrac{21}{48} = -\dfrac{7}{16}$. Hence

\[ e^{g(x)} = 1-\frac{x}{2}+\frac{11x^{2}}{24}-\frac{7x^{3}}{16}+O\!\left(x^{4}\right). \tag{4} \]

Step 4: Form the difference

\[ \bigl(1+x\bigr)^{1/x} = e-\frac{ex}{2}+\frac{11ex^{2}}{24}-\frac{7ex^{3}}{16}+O\!\left(x^{4}\right), \tag{5} \]

\[ \alpha(x) = e-\bigl(1+x\bigr)^{1/x} = \frac{ex}{2}-\frac{11ex^{2}}{24}+\frac{7ex^{3}}{16}+O\!\left(x^{4}\right). \tag{6} \]

Step 5: Determine the order

The leading term of (6) is $\dfrac{e}{2}x$, so test $n = 1$: \[\lim_{x\to0}\frac{e-\bigl(1+x\bigr)^{1/x}}{x} = \lim_{x\to0}\left(\frac{e}{2}-\frac{11e}{24}x+O\!\left(x^{2}\right)\right) = \frac{e}{2} \neq 0. \tag{7}\]

Step 6: Verification

\[\frac{e}{2} = 1.3591409:\quad x = 0.01:\ 1.3467999; \qquad x = 0.001:\ 1.3578962; \qquad x = -0.01:\ 1.3717198 \;\;\checkmark\]

\[\text{The gaps match } -\frac{11e}{24}x = -1.2459x:\ \text{at } x = 0.01,\ 1.3591-0.0125 = 1.3466 \;\;\checkmark\]

\[\text{Equivalently } \bigl(1+x\bigr)^{1/x} = e\left(1-\frac{x}{2}+\frac{11x^{2}}{24}-\cdots\right) \;\;\checkmark\]