If `alpha,beta` are the roots of the equation `ax^2+bx+c=0`, then `lim_(xrarralpha)(ax^2+bx+c+1)^(1//x-alpha) ` is equal to

Options

A. `2a(alpha-beta)`
B. `-2ln |a(alpha-beta)|`
C. `e^(a(alpha-beta))`
D. `e^(a^(2))|alpha-beta|`

Answer

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

### Step-by-Step Solution:

1. **Identify the Limit Expression**:
We need to evaluate the limit:
\[ L = \lim_{x \to \alpha} (ax^2 + bx + c + 1)^{\frac{1}{x - \alpha}} \]

2. **Substituting the Roots**:
Since \(\alpha\) is a root of the equation \(ax^2 + bx + c = 0\), we have:
\[ a\alpha^2 + b\alpha + c = 0 \]
Therefore, we can rewrite the expression inside the limit:
\[ ax^2 + bx + c + 1 = (ax^2 + bx + c) + 1 = 0 + 1 = 1 \quad \text{when } x = \alpha \]

3. **Form of the Limit**:
As \(x\) approaches \(\alpha\), the expression becomes:
\[ L = \lim_{x \to \alpha} (1)^{\frac{1}{x - \alpha}} = 1^{\infty} \]
This is an indeterminate form, so we can apply the logarithmic limit technique.

4. **Using the Exponential Limit**:
We can rewrite the limit using the exponential function:
\[ L = e^{\lim_{x \to \alpha} \left( (ax^2 + bx + c + 1) - 1 \right) \cdot \frac{1}{x - \alpha}} \]

5. **Simplifying the Expression**:
We know that:
\[ ax^2 + bx + c = 0 \implies ax^2 + bx + c + 1 - 1 = ax^2 + bx + c \]
Thus, we need to evaluate:
\[ L = e^{\lim_{x \to \alpha} (ax^2 + bx + c) \cdot \frac{1}{x - \alpha}} \]

6. **Factoring the Quadratic**:
Since \(ax^2 + bx + c\) can be factored using its roots \(\alpha\) and \(\beta\):
\[ ax^2 + bx + c = a(x - \alpha)(x - \beta) \]
Therefore:
\[ L = e^{\lim_{x \to \alpha} a(x - \alpha)(x - \beta) \cdot \frac{1}{x - \alpha}} = e^{\lim_{x \to \alpha} a(x - \beta)} \]

7. **Evaluating the Limit**:
Now, substituting \(x = \alpha\):
\[ L = e^{a(\alpha - \beta)} \]

### Final Result:
Thus, the limit evaluates to:
\[ L = e^{a(\alpha - \beta)} \]