Let a denote the base of the natural loganthim. The value of real number a for which the right hand limit
`lim_(x to )((1-x)^(1//x)-e^(-1))/(x^(e))` is equal to a nonzer real number is, _____

Answer

To solve the limit problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{(1 - x)^{\frac{1}{x}} - e^{-1}}{x^a} \]

where we want to find the value of \(a\) such that this limit is a non-zero real number.

### Step 1: Rewrite the limit
We start by rewriting the limit as:

\[ \lim_{x \to 0} \frac{(1 - x)^{\frac{1}{x}} - \frac{1}{e}}{x^a} \]

### Step 2: Analyze the expression \((1 - x)^{\frac{1}{x}}\)
As \(x\) approaches 0, we can use the fact that:

\[ (1 - x)^{\frac{1}{x}} \to e^{-1} \]

Thus, both the numerator and denominator approach 0, indicating that we have a \( \frac{0}{0} \) indeterminate form. This allows us to apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and denominator:

\[ \lim_{x \to 0} \frac{\frac{d}{dx}((1 - x)^{\frac{1}{x}} - e^{-1})}{\frac{d}{dx}(x^a)} \]

The derivative of \(e^{-1}\) is 0 since it is a constant.

### Step 4: Differentiate the numerator
Let \(y = (1 - x)^{\frac{1}{x}}\). Taking the natural logarithm:

\[ \ln y = \frac{1}{x} \ln(1 - x) \]

Differentiating both sides with respect to \(x\):

\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{x} \ln(1 - x)\right) \]

Using the product rule:

\[ \frac{d}{dx}\left(\frac{1}{x} \ln(1 - x)\right) = -\frac{1}{x^2} \ln(1 - x) + \frac{1}{x(1 - x)} \]

Thus,

\[ \frac{dy}{dx} = y \left(-\frac{1}{x^2} \ln(1 - x) + \frac{1}{x(1 - x)}\right) \]

### Step 5: Substitute back into the limit
Now substituting back into the limit gives:

\[ \lim_{x \to 0} \frac{(1 - x)^{\frac{1}{x}} \left(-\frac{1}{x^2} \ln(1 - x) + \frac{1}{x(1 - x)}\right)}{ax^{a-1}} \]

### Step 6: Simplify the limit
As \(x \to 0\), we know that:

\[ \ln(1 - x) \sim -x \quad \text{(using Taylor expansion)} \]

Thus, we can simplify our limit to:

\[ \lim_{x \to 0} \frac{e^{-1} \left(-\frac{1}{x^2}(-x) + \frac{1}{x}\right)}{ax^{a-1}} = \lim_{x \to 0} \frac{e^{-1} \left(\frac{1}{x} + \frac{1}{x}\right)}{ax^{a-1}} = \lim_{x \to 0} \frac{2e^{-1}}{ax^{a-2}} \]

### Step 7: Determine conditions for a non-zero limit
For the limit to be a non-zero real number, we need \(a - 2 = 0\), which implies:

\[ a = 2 \]

### Conclusion
Thus, the value of the real number \(a\) for which the limit is equal to a non-zero real number is:

\[ \boxed{2} \]