If `lim_(x to 1) (1 + ax + bx^(2))^((c )/(x - 1)) = e^(3)`, then the value of a + b + bc + 2009 is…….

Answer

To solve the limit problem given, we start with the expression:

\[ \lim_{x \to 1} \left(1 + ax + bx^2\right)^{\frac{c}{x - 1}} = e^3 \]

### Step 1: Rewrite the limit

We can rewrite the limit in a more manageable form. We know that if \( \lim_{x \to a} f(x) = 0 \), then we can use the exponential limit property:

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

Here, let \( f(x) = 1 + ax + bx^2 \) and \( g(x) = \frac{c}{x - 1} \).

### Step 2: Evaluate \( f(1) \)

First, we evaluate \( f(1) \):

\[ f(1) = 1 + a(1) + b(1^2) = 1 + a + b \]

For the limit to be valid, \( f(1) \) must equal 1, which gives us:

\[ 1 + a + b = 1 \implies a + b = 0 \]

### Step 3: Substitute \( a + b = 0 \)

Since \( a + b = 0 \), we can express \( b \) in terms of \( a \):

\[ b = -a \]

### Step 4: Rewrite \( f(x) \)

Now substitute \( b \) back into \( f(x) \):

\[ f(x) = 1 + ax - a x^2 = 1 + a(x - x^2) = 1 + a x(1 - x) \]

### Step 5: Find \( \lim_{x \to 1} f(x) \)

Next, we need to evaluate the limit:

\[ \lim_{x \to 1} f(x) = 1 + a(1)(1 - 1) = 1 \]

### Step 6: Evaluate \( \lim_{x \to 1} g(x) \)

Now we need to evaluate \( g(x) \):

\[ g(x) = \frac{c}{x - 1} \]

As \( x \to 1 \), \( g(x) \to \infty \).

### Step 7: Use L'Hôpital's Rule

We need to evaluate:

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

Using L'Hôpital's Rule since both the numerator and denominator approach 0:

\[ \lim_{x \to 1} \frac{f'(x)}{1} \]

### Step 8: Differentiate \( f(x) \)

Differentiate \( f(x) \):

\[ f'(x) = a + 2bx = a + 2(-a)x = a(1 - 2x) \]

Now evaluate at \( x = 1 \):

\[ f'(1) = a(1 - 2) = -a \]

### Step 9: Calculate the limit

Thus, we have:

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

### Step 10: Set up the equation

Now we can set up our limit equation:

\[ \lim_{x \to 1} f(x) g(x) = e^3 \implies c \cdot (-a) = 3 \implies c = -\frac{3}{a} \]

### Step 11: Substitute back

Now we need to find \( a + b + bc + 2009 \):

Substituting \( b = -a \):

\[ a + (-a) + (-a)\left(-\frac{3}{a}\right) + 2009 = 0 + 3 + 2009 = 2012 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{2012} \]