Find the order of the infinitesimal alpha(x) = ln(1+x)-e

Find the order of the infinitesimal $\alpha(x) = \ln(1+x)-e\ln\ln\bigl(e+x\bigr)$ with respect to $x$ as $x\to0$.

Step 1: Check that $\alpha$ is infinitesimal

\[ \alpha(0) = \ln1-e\ln\ln e = 0-e\ln1 = 0 \;\;\checkmark \tag{1} \]

Step 2: Expand the first term

\[ \ln(1+x) = x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+O\!\left(x^{4}\right). \tag{2} \]

Step 3: Expand the second term

Writing $e+x = e\left(1+\dfrac{x}{e}\right)$ gives $\ln\bigl(e+x\bigr) = 1+u$ with

\[ u = \ln\left(1+\frac{x}{e}\right) = \frac{x}{e}-\frac{x^{2}}{2e^{2}}+\frac{x^{3}}{3e^{3}}+O\!\left(x^{4}\right). \tag{3} \]

Then $\ln\ln\bigl(e+x\bigr) = \ln(1+u) = u-\dfrac{u^{2}}{2}+\dfrac{u^{3}}{3}+O\!\left(u^{4}\right)$, and with

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

the coefficients collect to $\dfrac1e$ for $x$; $-\dfrac{1}{2e^{2}}-\dfrac{1}{2e^{2}} = -\dfrac{1}{e^{2}}$ for $x^{2}$; and $\dfrac13+\dfrac12+\dfrac13 = \dfrac76$, divided by $e^{3}$, for $x^{3}$:

\[ \ln\ln\bigl(e+x\bigr) = \frac{x}{e}-\frac{x^{2}}{e^{2}}+\frac{7x^{3}}{6e^{3}}+O\!\left(x^{4}\right), \tag{5} \]

\[ e\ln\ln\bigl(e+x\bigr) = x-\frac{x^{2}}{e}+\frac{7x^{3}}{6e^{2}}+O\!\left(x^{4}\right). \tag{6} \]

Step 4: Subtract

The linear terms of (2) and (6) are both exactly $x$ and cancel: \[\alpha(x) = \left(\frac1e-\frac12\right)x^{2}+\left(\frac13-\frac{7}{6e^{2}}\right)x^{3}+O\!\left(x^{4}\right). \tag{7}\]

Step 5: Determine the order

Since $\dfrac1e-\dfrac12 = 0.3678794-0.5 = -0.1321206\neq0$, the test with $n = 2$ succeeds: \[\lim_{x\to0}\frac{\alpha(x)}{x^{2}} = \frac1e-\frac12 \neq 0. \tag{8}\]

Step 6: Verification

\[\frac{\alpha(x)}{x^{2}}:\quad x = 0.01:\ -0.1303837; \qquad x = 0.001:\ -0.1319453; \qquad x = -0.01:\ -0.1338929 \;\;\checkmark\]

\[\text{Target } \frac1e-\frac12 = -0.1321206;\ \text{the cubic coefficient } \frac13-\frac{7}{6e^{2}} = 0.1754422 \;\;\checkmark\]

\[\text{At } x = 0.01:\ -0.1321206+0.0017544 = -0.1303662 \;\;\checkmark\]