If `alpha, beta` are the roots of `x^(2)-ax+b=0`, then `lim_(xrarralpha)(e^(x^(2)-ax+b))/(x-alpha)=`

Options

A. `beta-alpha`
B. `alpha-beta`
C. 1
D. `2alpha-a`

Answer

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

\[ \lim_{x \to \alpha} \frac{e^{x^2 - ax + b}}{x - \alpha} \]

### Step 1: Substitute the expression for \(x^2 - ax + b\)

Since \(\alpha\) is a root of the quadratic equation \(x^2 - ax + b = 0\), we know that:

\[ \alpha^2 - a\alpha + b = 0 \]

This implies that:

\[ x^2 - ax + b = (x - \alpha)(x - \beta) \]

Thus, we can rewrite the limit as:

\[ \lim_{x \to \alpha} \frac{e^{(x - \alpha)(x - \beta)}}{x - \alpha} \]

### Step 2: Evaluate the limit

As \(x\) approaches \(\alpha\), the expression \((x - \alpha)(x - \beta)\) approaches \(0\). Therefore, we can rewrite the limit:

\[ \lim_{x \to \alpha} \frac{e^{(x - \alpha)(x - \beta)}}{x - \alpha} \]

This results in an indeterminate form of \(\frac{0}{0}\). To resolve this, we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

We differentiate the numerator and the denominator:

- The derivative of the numerator \(e^{(x - \alpha)(x - \beta)}\) using the chain rule is:

\[ e^{(x - \alpha)(x - \beta)} \cdot \frac{d}{dx}((x - \alpha)(x - \beta)) = e^{(x - \alpha)(x - \beta)} \cdot ((x - \beta) + (x - \alpha)) \]

- The derivative of the denominator \(x - \alpha\) is simply \(1\).

Thus, we have:

\[ \lim_{x \to \alpha} e^{(x - \alpha)(x - \beta)} \cdot ((x - \beta) + (x - \alpha)) \]

### Step 4: Evaluate the limit again

Now substituting \(x = \alpha\):

\[ = e^{0} \cdot ((\alpha - \beta) + 0) = 1 \cdot (\alpha - \beta) = \alpha - \beta \]

### Final Result

Thus, the limit evaluates to:

\[ \lim_{x \to \alpha} \frac{e^{x^2 - ax + b}}{x - \alpha} = \alpha - \beta \]