The value of the real number a for which the right hand limit $\underset{x\to 0^{+}}{lim} \frac{(1-x{)}^{1/x}-e^{-1}}{x^a}$ is equal to a non-zero real number, is

It is given that $\underset{x\to 0^{+}}{lim} \frac{(1-x{)}^{1/x}-e^{-1}}{x^a}$ exists and is a non-zero number

Now,

$\begin{array}{l}\underset{x\to 0^{+}}{lim} \frac{(1-x{)}^{1/x}-e^{-1}}{x^a}\\ =\underset{x\to 0^{+}}{lim} \frac{e^{\frac{1}{x}\log (1-x)}-e^{-1}}{x^a}=\underset{x\to 0^{+}}{lim} \frac{e^{-\left(1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\cdots \right)}-e^{-1}}{x^a}\end{array}$

$\begin{array}{l}=\underset{x\to 0^{+}}{lim} e^{-1}\left\{\frac{e^{-\left(\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\dots \right)}-1}{x^a}\right\}\\ =\underset{x\to 0^{+}}{lim} e^{-1}\left\{\frac{e^{-\left(\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\dots \right)}-1}{-\left(\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\dots \right)}\right\}\times -\frac{1}{x^a}\left(\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\dots \right)\\ =-e^{-1}\times 1\left\{\underset{x\to 0^{+}}{lim} \left(\frac{1}{2}{x}^{1-a}+\frac{1}{3}{x}^{2-a}+\frac{1}{4}{x}^{3-a}+\dots \right)\right\}\\ =-\frac{1}{e}\times \frac{1}{2},\text{ if }a=1.\end{array}$

Hence, the limit is a non-zero number when $a= 1.$