Question

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

Answer 1

Answer: 1

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