Question

Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit 

\( \lim\limits_{x \to 0^+} \frac{(1 - x)^\frac{1}{x}\,-\,e^{-1}}{x^a}\) 

is equal to a nonzero real number, is ______.

Answer 1

= \( \frac{1}{e}\lim\limits_{x - 0^+} \frac{(-x-\frac{x^2}{2}-\frac{x^3}{3}-...)+ x}{x^{a+1}}\)

Thus, a = 1