Let `f (x)` be a polynomial satisfying `lim _(x to oo) (x ^(4) f (x))/( x ^(8) +1)=3`
`f (2) =5, f(3) =10, f (-1)=2, f (-6)=37`
The value of `lim _(x to -6) (f (x) -x ^(2) -1)/(3 (x+6))` equals to:

Options

A. `-|underset-(6)`
B. `|underset-(6)`
C. `(|underset-(6))/(2)`
D. `(|underset-(-6))/(2)`

Answer

To solve the problem, we need to find the polynomial \( f(x) \) that satisfies the given conditions and then evaluate the limit. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the Degree of the Polynomial
We are given:
\[ \lim_{x \to \infty} \frac{x^4 f(x)}{x^8 + 1} = 3 \]
As \( x \to \infty \), the dominant term in the denominator is \( x^8 \). For the limit to exist and equal 3, the degree of \( f(x) \) must be 4. Thus, we can express \( f(x) \) as:
\[ f(x) = ax^4 + bx^3 + cx^2 + dx + e \]

### Step 2: Use Given Values to Formulate \( f(x) \)
We have the following values:
- \( f(2) = 5 \)
- \( f(3) = 10 \)
- \( f(-1) = 2 \)
- \( f(-6) = 37 \)

From these values, we can observe a pattern. It appears that:
- \( f(2) = 2^2 + 1 = 5 \)
- \( f(3) = 3^2 + 1 = 10 \)
- \( f(-1) = (-1)^2 + 1 = 2 \)
- \( f(-6) = (-6)^2 + 1 = 37 \)

This suggests that \( f(x) \) can be expressed as:
\[ f(x) = (x - 2)(x - 3)(x + 1)(x + 6) + x^2 + 1 \]
where the first part becomes zero at \( x = 2, 3, -1, -6 \) and the second part ensures the polynomial matches the values given.

### Step 3: Confirm the Leading Coefficient
To ensure that the leading coefficient of \( f(x) \) is consistent with the limit condition, we can expand the polynomial:
\[ f(x) = k(x - 2)(x - 3)(x + 1)(x + 6) + x^2 + 1 \]
The leading term of \( (x - 2)(x - 3)(x + 1)(x + 6) \) is \( x^4 \), and thus the leading term of \( f(x) \) will be \( kx^4 + x^2 + 1 \).

To satisfy the limit condition:
\[ \lim_{x \to \infty} \frac{x^4 (kx^4)}{x^8} = k = 3 \]
Thus, \( k = 3 \).

### Step 4: Write the Final Form of \( f(x) \)
Now we can write:
\[ f(x) = 3(x - 2)(x - 3)(x + 1)(x + 6) + x^2 + 1 \]

### Step 5: Evaluate the Limit
We need to find:
\[ \lim_{x \to -6} \frac{f(x) - x^2 - 1}{3(x + 6)} \]
Since both the numerator and denominator approach 0 as \( x \to -6 \), we can apply L'Hôpital's rule:
1. Differentiate the numerator: \( f'(x) - 2x \)
2. Differentiate the denominator: \( 3 \)

### Step 6: Calculate \( f'(-6) \)
To find \( f'(-6) \), we differentiate \( f(x) \):
Using the product rule:
\[ f'(x) = 3 \left[ (x - 3)(x + 1)(x + 6) + (x - 2)(x + 1)(x + 6) + (x - 2)(x - 3)(x + 6) + (x - 2)(x - 3)(x + 1) \right] \]
Now substitute \( x = -6 \) into \( f'(x) \).

### Step 7: Compute the Limit
Substituting back into the limit:
\[ \lim_{x \to -6} \frac{f'(x) - 2(-6)}{3} \]
This will yield the final result.

### Final Answer
After evaluating, we find that:
\[ \text{The value of } \lim_{x \to -6} \frac{f(x) - x^2 - 1}{3(x + 6)} = -\frac{720}{2} = -360 \]