If `lim_(xto1) (1+ax+bx^(2))^((e)/((x-1)))=e^(3)`, then the value of bc is _________.

Options

A. `a=2`
B. `a=1`
C. `L=(1)/(64)`
D. L=(1)/(32)`

Explanation

To solve the limit problem given in the question, we can follow these steps:

### Step 1: Identify the limit expression
We start with the limit:
\[ \lim_{x \to 1} (1 + ax + bx^2)^{\frac{e}{x-1}} = e^3 \]

### Step 2: Recognize the indeterminate form
As \(x\) approaches 1, the expression \(1 + ax + bx^2\) approaches \(1 + a + b\). Thus, we have the form \(1^\infty\) which is an indeterminate form.

### Step 3: Apply the logarithmic limit property
We can rewrite the limit using the property of logarithms:
\[ \lim_{x \to 1} (1 + ax + bx^2)^{\frac{e}{x-1}} = e^{\lim_{x \to 1} \frac{e}{x-1} \ln(1 + ax + bx^2)} \]
Setting this equal to \(e^3\), we have:
\[ \lim_{x \to 1} \frac{e}{x-1} \ln(1 + ax + bx^2) = 3 \]

### Step 4: Simplify the logarithmic expression
Using the Taylor expansion for \(\ln(1 + u)\) around \(u = 0\):
\[ \ln(1 + u) \approx u \quad \text{for small } u \]
We can express:
\[ \ln(1 + ax + bx^2) \approx ax + bx^2 \quad \text{as } x \to 1 \]
Thus, we have:
\[ \ln(1 + ax + bx^2) \approx a + b \quad \text{when } x \to 1 \]

### Step 5: Set up the limit
Now substituting back into our limit:
\[ \lim_{x \to 1} \frac{e}{x-1} (a + b) = 3 \]
This implies:
\[ e(a + b) = 3 \quad \text{(as } \lim_{x \to 1} \frac{1}{x-1} \text{ diverges)} \]

### Step 6: Solve for \(a + b\)
From this, we can deduce:
\[ a + b = \frac{3}{e} \]

### Step 7: Use the condition for the limit
We also know that for the limit to exist, the expression \(1 + ax + bx^2\) must approach 1 as \(x \to 1\). Therefore:
\[ a + b = 0 \quad \text{(for the limit to be valid)} \]
This gives us our first equation:
1. \(a + b = 0\)

### Step 8: Solve for \(b\) in terms of \(a\)
From \(a + b = 0\), we can express \(b\) as:
\[ b = -a \]

### Step 9: Substitute into the second equation
Substituting \(b = -a\) into the equation \(a + b = \frac{3}{e}\):
\[ a - a = \frac{3}{e} \]
This is a contradiction unless we consider the limit behavior.

### Step 10: Analyze the coefficients
Now we need to analyze the coefficients of the limit expression:
Using L'Hôpital's Rule on:
\[ \lim_{x \to 1} \frac{(ax + bx^2)}{(x - 1)} \]
We differentiate the numerator and denominator:
\[ \lim_{x \to 1} \frac{a + 2bx}{1} = a + 2b \]

### Step 11: Set the limit equal to 3
Setting this equal to 3 gives us:
\[ a + 2b = 3 \]

### Step 12: Solve the system of equations
Now we have a system of equations:
1. \(a + b = 0\)
2. \(a + 2b = 3\)

Substituting \(b = -a\) into the second equation:
\[ a + 2(-a) = 3 \implies -a = 3 \implies a = -3 \]
Then substituting back:
\[ b = -(-3) = 3 \]

### Step 13: Calculate \(bc\)
Now we need to find \(bc\):
\[ bc = (-3)(3) = -9 \]

### Final Answer
The value of \(bc\) is:
\[ \boxed{-9} \]