If f(x) is a differentiable function such that f'(1) = 4 and f'(4) =`1/2` then value of `lim_(xrarr0)(f(x^2+x+1)-f(1))/(f(x^(4)-x^2+2x+4)-f(4))`

Options

A. 8
B. 16
C. 4
D. Does not exist

Answer

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

\[ \lim_{x \to 0} \frac{f(x^2 + x + 1) - f(1)}{f(x^4 - x^2 + 2x + 4) - f(4)} \]

### Step 1: Evaluate the limit as \( x \to 0 \)

First, we substitute \( x = 0 \) into the expression:

- For the numerator:
\[ f(0^2 + 0 + 1) - f(1) = f(1) - f(1) = 0 \]

- For the denominator:
\[ f(0^4 - 0^2 + 2 \cdot 0 + 4) - f(4) = f(4) - f(4) = 0 \]

Since both the numerator and denominator approach 0, we have a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to \( x \):

- **Numerator**:
\[ \text{Let } u = x^2 + x + 1 \Rightarrow \frac{du}{dx} = 2x + 1 \]
Thus, the derivative of the numerator is:
\[ f'(u) \cdot (2x + 1) = f'(x^2 + x + 1)(2x + 1) \]

- **Denominator**:
\[ \text{Let } v = x^4 - x^2 + 2x + 4 \Rightarrow \frac{dv}{dx} = 4x^3 - 2x + 2 \]
Thus, the derivative of the denominator is:
\[ f'(v) \cdot (4x^3 - 2x + 2) = f'(x^4 - x^2 + 2x + 4)(4x^3 - 2x + 2) \]

### Step 3: Rewrite the limit using derivatives

Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{f'(x^2 + x + 1)(2x + 1)}{f'(x^4 - x^2 + 2x + 4)(4x^3 - 2x + 2)} \]

### Step 4: Evaluate the limit again as \( x \to 0 \)

Substituting \( x = 0 \) into the derivatives:

- For the numerator:
\[ f'(1)(2 \cdot 0 + 1) = f'(1) \cdot 1 = f'(1) \]

- For the denominator:
\[ f'(4)(4 \cdot 0^3 - 2 \cdot 0 + 2) = f'(4) \cdot 2 \]

Thus, the limit simplifies to:

\[ \lim_{x \to 0} \frac{f'(1)}{2f'(4)} \]

### Step 5: Substitute the known values

We know from the problem statement that:

- \( f'(1) = 4 \)
- \( f'(4) = \frac{1}{2} \)

Substituting these values into the limit gives:

\[ \frac{f'(1)}{2f'(4)} = \frac{4}{2 \cdot \frac{1}{2}} = \frac{4}{1} = 4 \]

### Final Answer

Thus, the value of the limit is:

\[ \boxed{4} \]