Question

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

Answer 1

$\displaystyle\lim _{x \rightarrow 0^{+}} \frac{(1-x)^{1 / x}-\frac{1}{e}}{x^{a}}=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{e^{\frac{\ln (1-x)}{x}}-\frac{1}{e}}{x^{a}}$
$=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{1}{e} \frac{\left(e^{\frac{-x-\frac{x^{2}}{2}}{x}}-1\right)}{x^{a}}=\displaystyle\lim _{x \rightarrow 0} \frac{1}{e}\left(\frac{e^{-\frac{x}{2}}-1}{x^{a}}\right)$
For value of limit to be a non-zero real number $a =1$