\(\lim\limits_{x\to 0^+} \cfrac{(1 - x)^\frac 1x - e^{-1}}{x^a}\) \(\left(\frac 00 -case\right)\)
\(= \lim\limits_{x+0^+} \cfrac{\left( - \frac{log(1-x)}{x^2} - \frac1{x(1 -x)}\right)(1-x)^\frac1x}{ax^{a-1}}\) (By using D.L.H. Rule)
\(= \lim\limits_{x\to 0^+} \cfrac{-((1-x)log(1-x) + x)(1-x)^\frac 1x}{x^2(1-x). ax^{a-1}}\)
\(= -\frac{e^{-1}}a \lim\limits_{x\to 0^+} \cfrac{(1 - x)log(1-x) + x}{x^{a + 1}}\) \(\left(\frac 00 -case\right)\)
\(= -\frac{e^{-1}}a \lim\limits_{x\to 0^+} \cfrac{(1 - x)(-x-\frac{x^2}2 - \frac{x^3}3 - ...) + x}{x^{a + 1}}\)
\(= \frac{e^{-1}}a \lim\limits_{x\to 0^+} \cfrac{(x+\frac{x^2}2 +\frac{x^3}3+ ...) -(x^2+\frac{x^3}2 +\frac{x^4}3 + ...)- x}{x^{a + 1}}\)
\(= \frac{e^{-1}}a \lim\limits_{x\to 0^+} \cfrac{-\frac{x^2}2 - \frac{x^3}6 - ....}{x^{a +1}}\)
For limit to be exist a+1 should be equal to 2.
\(\therefore a + 1 = 2\)
⇒ \(a = 1\)
