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

Options

A. `e^(a(alpha-beta))`
B. `e^(a(beta-alpha))`
C. 1
D. none of these

Explanation

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

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

### Step 1: Rewrite the quadratic expression
Since \(\alpha\) and \(\beta\) are roots of the quadratic equation \(ax^2 + bx + c = 0\), we can express the quadratic as:

\[ ax^2 + bx + c = a(x - \alpha)(x - \beta) \]

### Step 2: Substitute the quadratic expression into the limit
We substitute this expression into our limit:

\[ \lim_{x \to \alpha} \left( 1 + a(x - \alpha)(x - \beta) \right)^{\frac{1}{x - \alpha}} \]

### Step 3: Simplify the expression inside the limit
As \(x\) approaches \(\alpha\), the term \(a(x - \alpha)(x - \beta)\) approaches \(0\). Therefore, we can rewrite the limit as:

\[ \lim_{x \to \alpha} \left( 1 + 0 \right)^{\frac{1}{x - \alpha}} = \lim_{x \to \alpha} 1^{\frac{1}{x - \alpha}} = 1 \]

However, this is an indeterminate form of type \(1^{\infty}\). To resolve this, we can use the exponential limit formula.

### Step 4: Use the exponential limit formula
We apply the formula:

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

Here, \(f(x) = 1 + a(x - \alpha)(x - \beta)\) and \(g(x) = \frac{1}{x - \alpha}\).

### Step 5: Calculate \(f(x) - 1\)
We find:

\[ f(x) - 1 = a(x - \alpha)(x - \beta) \]

### Step 6: Calculate the limit
Now, we compute:

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

### Step 7: Final result
Thus, we have:

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

So, the final answer is:

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