Let e denote the base of the natural logarithm. The value of the...

$\lim_{x \to {0}^{+}}\dfrac{{\left(1 - x\right)}^{\dfrac{1}{x}}}{a{x}^{a - 1}}\dfrac{d}{dx}\left[\dfrac{\ln \left(1 - x\right)}{x}\right] = \lim_{x \to {0}^{+}}\dfrac{\left(1 - x\right)\dfrac{1}{x}}{a{x}^{a - 1}}\left[\dfrac{1}{\left(x\right)\left(x - 1\right)} - \dfrac{\ln \left(1 - x\right)}{{x}^{2}}\right]$

$= \lim_{x \to {0}^{+}}\dfrac{{\left(1 - x\right)}^{\dfrac{1}{x}}}{a{x}^{a - 1}}\left[\dfrac{x - \left(x - 1\right)\ln \left(1 - x\right)}{{x}^{2}\left(x - 1\right)}\right]$

$= \lim_{x \to {0}^{+}}\dfrac{\left(1 - x\right)\dfrac{1}{x}\left\{x + \left(x - 1\right)\left(x + \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{3} . . . . . . . .\right)\right\}}{- a\left(1 - x\right){x}^{a + 1}}$

$= \lim_{x \to {0}^{+}}\dfrac{\left(1 - x\right)\dfrac{1}{x}\left\{\dfrac{{x}^{2}}{2} + {x}^{3}\left(\dfrac{1}{2} - \dfrac{1}{3}\right) + {x}^{4}\left(\dfrac{1}{3} - \dfrac{1}{4}\right) + \ldots\right\}}{- a\left(1 - x\right){x}^{a + 1}}$

Which is real number if and only if $a + 1 = 2 ⇒ a = 1$