Evaluate L = lim as x to +0 x^(1/ln (1-e^-x))

Evaluate $\displaystyle L = \lim_{x\to+0}x^{\,1/\ln\left(1-e^{-x}\right)}$.

Step 1: Take the natural logarithm

As $x\to+0$ the base tends to $0$ and $\ln\bigl(1-e^{-x}\bigr)\to-\infty$, so the exponent tends to $0$: the form is $0^{0}$. Write $L = e^{I}$ with

\[ I = \lim_{x\to+0}\frac{\ln x}{\ln\bigl(1-e^{-x}\bigr)}. \]

Step 2: Expand the denominator

Since $1-e^{-x} = \dfrac{e^{x}-1}{e^{x}}$, and $e^{x}-1 = x+\dfrac{x^{2}}{2}+O(x^{3})$, \[1-e^{-x} = \frac{x+\dfrac{x^{2}}{2}+O\!\left(x^{3}\right)}{1+x+O\!\left(x^{2}\right)} = x\cdot\frac{1+\dfrac{x}{2}+O\!\left(x^{2}\right)}{1+x+O\!\left(x^{2}\right)}.\]

Taking logarithms and using $\ln(1+u) = u+O(u^{2})$: \[\ln\bigl(1-e^{-x}\bigr) = \ln x+\frac{x}{2}-x+O\!\left(x^{2}\right) = \ln x-\frac{x}{2}+O\!\left(x^{2}\right). \tag{1}\]

Step 3: Form the ratio

Dividing numerator and denominator by $\ln x$: \[I = \lim_{x\to+0}\frac{\ln x}{\ln x-\dfrac{x}{2}+O\!\left(x^{2}\right)} = \lim_{x\to+0}\frac{1}{1-\dfrac{x}{2\ln x}+O\!\left(\dfrac{x^{2}}{\ln x}\right)}. \tag{2}\]

Here $\dfrac{x}{\ln x}\to0$, because $x\to0$ while $|\ln x|\to\infty$, so every correction disappears: \[I = \frac{1}{1-0} = 1. \tag{3}\]

Step 4: Recover $L$

\[ L = e^{I} = e. \]

Step 5: Verification

\[1-e^{-x}\sim x \;\Longrightarrow\; \frac{\ln x}{\ln\bigl(1-e^{-x}\bigr)}\sim\frac{\ln x}{\ln x} = 1 \;\;\checkmark\]

\[x = 0.01:\ 2.7153377; \qquad x = 0.001:\ 2.7180851; \qquad x = 10^{-6}:\ 2.7182817 \;\;\checkmark\]

\[\text{The relative gap behaves like } \frac{x}{2\ln x}:\ \text{at } x = 0.01,\ -1.1\cdot10^{-3} \;\;\checkmark\]