Step-by-Step Solution

1. Analyze the Limit Expression We are given the limit: $L = \mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$ The problem states that this limit L is a non-zero real number, and we need to find the value of a.

2. Evaluate the Numerator as x approaches 0 Let's first find the limit of the term $(1 - x)^(1/x)$ as $x -> 0^+$. This is a standard limit form related to the definition of e. $\mathop {\lim }\limits_{x \to 0} {(1 - x)^{1/x}} = \mathop {\lim }\limits_{x \to 0} {\left[ {{{(1 + ( - x))}^{1/( - x)}}} \right]^{ - 1}}$ Using the standard limit $\mathop {\lim }\limits_{t \to 0} {(1 + t)^{1/t}} = e$, with t = -x, we get: $\mathop {\lim }\limits_{x \to 0} {(1 - x)^{1/x}} = {e^{ - 1}}$ Therefore, as $x -> 0^+$, the numerator $(1 - x)^(1/x) - e^(-1)$ approaches $e^(-1) - e^(-1) = 0$.

3. Analyze the Denominator and the Form of the Limit As $x -> 0^+$, the denominator $x^a$ approaches 0 if a > 0.

   • If a = 0, the denominator is 1, and the limit becomes 0/1 = 0, which contradicts the condition that the limit is non-zero.
   • If a < 0, let a = -b where b > 0. The denominator $x^(-b) = 1/x^b$ approaches +∞. The limit becomes 0/∞ = 0, which is also a contradiction. So, we must have a > 0. This means the limit is of the indeterminate form 0/0.

4. Use Taylor Series Expansion for the Numerator To resolve the 0/0 form and find the value of a, we can use Taylor series expansion for the term $(1 - x)^(1/x)$ around x = 0.

   First, rewrite the term using the exponential function: ${(1 - x)^{1/x}} = {e^{\frac{1}{x}\ln (1 - x)}}$

   Now, use the Maclaurin series for ln(1 - u) with u = x: $\ln (1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \dots$

   Substitute this into the exponent: $\frac{1}{x}\ln (1 - x) = \frac{1}{x}\left( { - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \dots } \right) = - 1 - \frac{x}{2} - \frac{{{x^2}}}{3} - \dots$

   Now, substitute this back into the exponential expression: ${(1 - x)^{1/x}} = {e^{ - 1 - \frac{x}{2} - \frac{{{x^2}}}{3} - \dots }} = {e^{ - 1}} \cdot {e^{ - \frac{x}{2} - \frac{{{x^2}}}{3} - \dots }}$

   Next, use the Maclaurin series for $e^u$ with $u = -x/2 - x^2/3 - ...$. As x -> 0, u -> 0. ${e^u} = 1 + u + \frac{{{u^2}}}{{2!}} + \dots$ ${e^{ - \frac{x}{2} - \frac{{{x^2}}}{3} - \dots }} = 1 + \left( { - \frac{x}{2} - \frac{{{x^2}}}{3} - \dots } \right) + O({x^2}) = 1 - \frac{x}{2} + O({x^2})$

   Combining these results: ${(1 - x)^{1/x}} = {e^{ - 1}}\left( {1 - \frac{x}{2} + O({x^2})} \right) = {e^{ - 1}} - \frac{{{e^{ - 1}}}}{2}x + O({x^2})$

5. Substitute the Expansion into the Limit Now, substitute this expansion back into the original limit expression: $L = \mathop {\lim }\limits_{x \to {0^ + }} {{\left( {{e^{ - 1}} - \frac{{{e^{ - 1}}}}{2}x + O({x^2})} \right) - {e^{ - 1}}} \over {{x^a}}}$ $L = \mathop {\lim }\limits_{x \to {0^ + }} {{ - \frac{{{e^{ - 1}}}}{2}x + O({x^2})} \over {{x^a}}}$ $L = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - \frac{{{e^{ - 1}}}}{2}{x^{1 - a}} + O({x^{2 - a}})} \right)$

6. Determine the Value of a For the limit L to be a non-zero real number, the exponent of the dominant term (the term with the lowest power of x) must be zero. The dominant term is $- (e^(-1)/2) * x^(1 - a)$. Thus, we must have: $1 - a = 0 \implies a = 1$

7. Verify the Result If a = 1, the limit becomes: $L = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - \frac{{{e^{ - 1}}}}{2}{x^{1 - 1}} + O({x^{2 - 1}})} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - \frac{{{e^{ - 1}}}}{2}{x^0} + O(x)} \right)$ $L = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - \frac{{{e^{ - 1}}}}{2} + O(x)} \right) = - \frac{{{e^{ - 1}}}}{2} = - \frac{1}{{2e}}$ Since -1/(2e) is a non-zero real number, our value a = 1 is correct.

Conclusion

The value of the real number a for which the limit is a non-zero real number is 1.