Evaluate L = lim as x to 1 x^(1/(1-x))

Evaluate $\displaystyle L = \lim_{x\to1}x^{1/(1-x)}$.

Step 1: Take the natural logarithm

The base tends to $1$ and the exponent $\dfrac{1}{1-x}$ becomes infinite: the form is $1^{\infty}$. Write $L = e^{I}$ with

\[ I = \lim_{x\to1}\frac{\ln x}{1-x}. \]

At $x = 1$ both $\ln1 = 0$ and $1-1 = 0$, so this is $\dfrac00$.

Step 2: Apply l'Hôpital's rule

\[ I = \lim_{x\to1}\frac{\dfrac{d}{dx}\ln x}{\dfrac{d}{dx}\bigl(1-x\bigr)} = \lim_{x\to1}\frac{1/x}{-1} = \lim_{x\to1}\left(-\frac1x\right). \tag{1} \]

Step 3: Evaluate the exponent

\[ I = -\frac11 = -1. \tag{2} \]

Step 4: Recover $L$

\[ L = e^{I} = e^{-1} = \frac1e. \]

Step 5: Verification

\[x = 0.999:\ 0.3676954; \qquad x = 1.001:\ 0.3680633; \qquad \tfrac1e = 0.3678794 \;\;\checkmark\]

\[\text{Substituting } x = 1+t:\ \frac{\ln(1+t)}{-t} = \frac{t-\tfrac{t^{2}}{2}+\cdots}{-t} = -1+\frac{t}{2}+\cdots \to -1 \;\;\checkmark\]

\[\text{Equivalently } x^{1/(1-x)} = \bigl(1+t\bigr)^{-1/t} \to e^{-1},\ \text{the classical limit} \;\;\checkmark\]