For a differentiable function `f(x)`, if `f'(2)=2 and f'(3)=1`, then the value of `lim_(xrarr0)(f(x^(2)+x+2)-f(2))/(f(x^(2)-x+3)-f(3))` is equal to

Options

A. 2
B. 1
C. `-2`
D. `-1`

Answer

To solve the limit problem step by step, we will use the information given about the differentiable function \( f(x) \) and apply L'Hôpital's Rule since we encounter a \( \frac{0}{0} \) form.

### Step 1: Identify the limit expression
We need to evaluate the limit:
\[ \lim_{x \to 0} \frac{f(x^2 + x + 2) - f(2)}{f(x^2 - x + 3) - f(3)} \]

### Step 2: Substitute \( x = 0 \)
Substituting \( x = 0 \) in the limit expression:
- The numerator becomes \( f(0^2 + 0 + 2) - f(2) = f(2) - f(2) = 0 \).
- The denominator becomes \( f(0^2 - 0 + 3) - f(3) = f(3) - f(3) = 0 \).

Thus, we have a \( \frac{0}{0} \) indeterminate form.

### Step 3: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

The limit now becomes:
\[ \lim_{x \to 0} \frac{f'(x^2 + x + 2) \cdot (2x + 1)}{f'(x^2 - x + 3) \cdot (2x - 1)} \]

### Step 4: Substitute \( x = 0 \) again
Now we substitute \( x = 0 \) into the new limit expression:
- The numerator becomes \( f'(0^2 + 0 + 2) \cdot (2 \cdot 0 + 1) = f'(2) \cdot 1 = f'(2) \).
- The denominator becomes \( f'(0^2 - 0 + 3) \cdot (2 \cdot 0 - 1) = f'(3) \cdot (-1) = -f'(3) \).

### Step 5: Use the given values of derivatives
We know from the problem statement:
- \( f'(2) = 2 \)
- \( f'(3) = 1 \)

Substituting these values into the limit gives:
\[ \lim_{x \to 0} \frac{f'(2)}{-f'(3)} = \frac{2}{-1} = -2 \]

### Conclusion
Thus, the value of the limit is:
\[ \boxed{-2} \]