Let f(x) is a polynomial function and `f(alpha))^(2) + f'(alpha))^(2) = 0`, then find `lim_(x rarr alpha) (f(x))/(f'(x))[(f'(x))/(f(x))]`, where [.] denotes greatest integer function, is........

Answer

To solve the problem, we start with the given condition:

1. **Given Condition**:
\[ f(\alpha)^2 + f'(\alpha)^2 = 0 \]
Since both \( f(\alpha)^2 \) and \( f'(\alpha)^2 \) are non-negative, the only way their sum can equal zero is if both terms are zero:
\[ f(\alpha) = 0 \quad \text{and} \quad f'(\alpha) = 0 \]

2. **Roots of the Polynomial**:
Since \( f(\alpha) = 0 \) and \( f'(\alpha) = 0 \), we can conclude that \( \alpha \) is a double root of the polynomial \( f(x) \). Thus, we can express \( f(x) \) as:
\[ f(x) = (x - \alpha)^2 g(x) \]
where \( g(x) \) is some polynomial function.

3. **Finding the Derivative**:
We differentiate \( f(x) \):
\[ f'(x) = 2(x - \alpha)g(x) + (x - \alpha)^2 g'(x) \]

4. **Evaluating the Limit**:
We need to find:
\[ \lim_{x \to \alpha} \frac{f(x)}{f'(x)} \left\lfloor \frac{f'(x)}{f(x)} \right\rfloor \]

5. **Calculating \( \frac{f(x)}{f'(x)} \)**:
Substituting \( f(x) \) and \( f'(x) \):
\[ f(x) = (x - \alpha)^2 g(x) \]
\[ f'(\alpha) = 2(x - \alpha)g(x) + (x - \alpha)^2 g'(\alpha) \]
At \( x = \alpha \):
\[ f'(\alpha) = 0 \quad \text{(since \( f'(\alpha) = 0 \))} \]
Therefore, we can rewrite:
\[ f'(x) \approx 2(x - \alpha)g(\alpha) \quad \text{(as \( x \to \alpha \))} \]
Thus,
\[ \frac{f(x)}{f'(x)} = \frac{(x - \alpha)^2 g(x)}{2(x - \alpha)g(\alpha)} = \frac{(x - \alpha) g(x)}{2g(\alpha)} \]

6. **Taking the Limit**:
As \( x \to \alpha \):
\[ \lim_{x \to \alpha} \frac{(x - \alpha) g(x)}{2g(\alpha)} = \frac{0 \cdot g(\alpha)}{2g(\alpha)} = 0 \]

7. **Calculating \( \frac{f'(x)}{f(x)} \)**:
We have:
\[ \frac{f'(x)}{f(x)} = \frac{2(x - \alpha)g(x) + (x - \alpha)^2 g'(x)}{(x - \alpha)^2 g(x)} = \frac{2}{x - \alpha} + \frac{g'(x)}{g(x)} \]
As \( x \to \alpha \), \( \frac{2}{x - \alpha} \to \infty \) and \( \frac{g'(x)}{g(x)} \) approaches a finite limit.

8. **Final Limit Calculation**:
Since \( \frac{f'(x)}{f(x)} \to \infty \), the greatest integer function \( \left\lfloor \frac{f'(x)}{f(x)} \right\rfloor \) approaches infinity. Therefore:
\[ \lim_{x \to \alpha} \frac{f(x)}{f'(x)} \left\lfloor \frac{f'(x)}{f(x)} \right\rfloor = 0 \cdot \infty = 0 \]

Thus, the final answer is:
\[ \boxed{1} \]