If a `lim_(x to 1) x^(1//(1-x)) + b =e^(-1)` where `a ge 1` and `b ge 0` then (a,b) is equal to:

Options

A. `(1, e^(-1))`
B. `(2, e^(-1))`
C. (1,0)
D. none of these

Answer

To solve the problem, we need to evaluate the limit and find the values of \(a\) and \(b\) such that:

\[ \lim_{x \to 1} a \cdot x^{\frac{1}{1-x}} + b = e^{-1} \]

### Step 1: Evaluate the limit

We start by evaluating the limit \( \lim_{x \to 1} x^{\frac{1}{1-x}} \). As \(x\) approaches 1, the expression \(x^{\frac{1}{1-x}}\) takes the form \(1^{\infty}\), which is an indeterminate form. We can use the following limit property:

\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} (f(x) - 1) \cdot g(x)} \]

Here, let \(f(x) = x\) and \(g(x) = \frac{1}{1-x}\). Thus, we can rewrite the limit as:

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

### Step 2: Simplify the limit

Now, we simplify the limit:

\[ \lim_{x \to 1} (x - 1) \cdot \frac{1}{1-x} = \lim_{x \to 1} \frac{x - 1}{1 - x} = \lim_{x \to 1} -1 = -1 \]

Thus, we have:

\[ \lim_{x \to 1} x^{\frac{1}{1-x}} = e^{-1} \]

### Step 3: Substitute back into the equation

Now we substitute this result back into our original equation:

\[ \lim_{x \to 1} a \cdot x^{\frac{1}{1-x}} + b = a \cdot e^{-1} + b \]

Setting this equal to \(e^{-1}\):

\[ a \cdot e^{-1} + b = e^{-1} \]

### Step 4: Rearranging the equation

Rearranging gives us:

\[ a \cdot e^{-1} + b = e^{-1} \implies a \cdot e^{-1} = e^{-1} - b \]

### Step 5: Solve for \(a\) and \(b\)

From the equation \(a \cdot e^{-1} = e^{-1} - b\), we can express \(b\):

\[ b = e^{-1} - a \cdot e^{-1} = e^{-1}(1 - a) \]

### Step 6: Apply the constraints

Given that \(a \geq 1\) and \(b \geq 0\):

1. Since \(b \geq 0\), we have:
\[ e^{-1}(1 - a) \geq 0 \implies 1 - a \geq 0 \implies a \leq 1 \]

2. Combining \(a \geq 1\) and \(a \leq 1\) gives us \(a = 1\).

### Step 7: Find \(b\)

Substituting \(a = 1\) back into the equation for \(b\):

\[ b = e^{-1}(1 - 1) = 0 \]

### Final Result

Thus, we find:

\[ (a, b) = (1, 0) \]