Parameter Value for Nonzero Right Hand Limit

To determine the value of the parameter $a$ for which the right-hand limit $\lim_{x \to 0^+} \frac{(1 - x)^{\frac{1}{x}} - e^{-1}}{x^a}$ is a non-zero real number, we expand the numerator using Taylor series around $x = 0$.

Using the expansion $\ln(1-x) = -x - \frac{x^2}{2} + O(x^3)$, we get: $(1 - x)^{\frac{1}{x}} = \exp\left(\frac{\ln(1-x)}{x}\right) = \exp\left(-1 - \frac{x}{2} + O(x^2)\right) = e^{-1} - \frac{e^{-1}}{2}x + O(x^2)$

Substituting this back into the limit yields: $\lim_{x \to 0^+} \frac{-\frac{e^{-1}}{2}x + O(x^2)}{x^a} = -\frac{e^{-1}}{2} \lim_{x \to 0^+} x^{1-a}$

For this limit to be a non-zero real number, the power of $x$ must be zero, which implies $1 - a = 0$.

Thus, the value of $a$ is $1$.