Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right hand limit $\lim_{x\to 0^{+}} \frac{(1-x)^{\frac{1}{x}} - e^{-1}}{x^{a}}$ is equal to a non-zero real number, is ___.

$\lim_{x\to 0^{+}} \dfrac{(1-x)^{\frac{1}{x}} - e^{-1}}{x^{a}}$

$e = \text{base of natural log},\quad a \in \mathbb{R}$

$\text{To find: } a \text{ such that the limit is non-zero and finite}$

$\lim_{x\to0^+}(1-x)^{1/x} = e^{-1}$ $\text{form } \dfrac{0}{0}$ $1.\ (1-x)^{1/x} = e^{\frac1x\ln(1-x)}$ $2.\ \ln(1-x) \text{ series}$ $3.\ \text{lowest power of } x \Rightarrow a$ $\ln(1-x) = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} - \cdots$ $\dfrac1x\ln(1-x) = -1 - \dfrac{x}{2} - \dfrac{x^2}{3} - \cdots$ $(1-x)^{1/x} = e^{-1}\,e^{-x/2 - x^2/3 - \cdots}$ $= e^{-1}\left[1 - \dfrac{x}{2} + \left(\dfrac18-\dfrac13\right)x^2 + \cdots\right]$ $= e^{-1}\left[1 - \dfrac{x}{2} - \dfrac{5x^2}{24} + \cdots\right]$ $(1-x)^{1/x} - e^{-1} = -e^{-1}\left[\dfrac{x}{2} + \dfrac{5x^2}{24} + \cdots\right]$ $\dfrac{(1-x)^{1/x}-e^{-1}}{x^{a}} = -\dfrac{e^{-1}}{2}\,x^{1-a} + O(x^{2-a})$ $1-a = 0 \Rightarrow a = 1$ $\lim_{x\to0^+} = -\dfrac{e^{-1}}{2} = -\dfrac{1}{2e} \neq 0$ $a<1:\ \to 0;\qquad a>1:\ \to \infty\ (\text{not finite non-zero})$ $\boxed{a=1}$