`f(x)` is a differentiable functions given `f'(1) = 4, f'(2) = 6`, where `f'(c )` means the derivatives of function at x = c, then
`lim_(h to 0) (f(2+2h + h^(2))-f(2))/(f(1+h-h^(2))-f(1))`

Options

A. not exist
B. `-3`
C. 3
D. `3//2`

Answer

To solve the limit

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

we start by recognizing that as \( h \to 0 \), both the numerator and the denominator approach 0, leading to the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator.

### Step 1: Differentiate the Numerator
The numerator is \( f(2 + 2h + h^2) - f(2) \). We differentiate it with respect to \( h \):

Using the chain rule, we have:

\[ \frac{d}{dh}[f(2 + 2h + h^2)] = f'(2 + 2h + h^2) \cdot \frac{d}{dh}(2 + 2h + h^2) = f'(2 + 2h + h^2) \cdot (2 + 2h). \]

### Step 2: Differentiate the Denominator
The denominator is \( f(1 + h - h^2) - f(1) \). We differentiate it similarly:

Using the chain rule again, we have:

\[ \frac{d}{dh}[f(1 + h - h^2)] = f'(1 + h - h^2) \cdot \frac{d}{dh}(1 + h - h^2) = f'(1 + h - h^2) \cdot (1 - 2h). \]

### Step 3: Apply L'Hôpital's Rule
Now we apply L'Hôpital's Rule:

\[ \lim_{h \to 0} \frac{f(2 + 2h + h^2) - f(2)}{f(1 + h - h^2) - f(1)} = \lim_{h \to 0} \frac{f'(2 + 2h + h^2) \cdot (2 + 2h)}{f'(1 + h - h^2) \cdot (1 - 2h)}. \]

### Step 4: Evaluate the Limit
As \( h \to 0 \):

- \( 2 + 2h \to 2 \)
- \( 1 + h - h^2 \to 1 \)
- \( f'(2 + 2h + h^2) \to f'(2) = 6 \)
- \( f'(1 + h - h^2) \to f'(1) = 4 \)

Thus, we have:

\[ \lim_{h \to 0} \frac{f'(2 + 2h + h^2) \cdot (2 + 2h)}{f'(1 + h - h^2) \cdot (1 - 2h)} = \frac{f'(2) \cdot 2}{f'(1) \cdot 1} = \frac{6 \cdot 2}{4 \cdot 1} = \frac{12}{4} = 3. \]

### Final Answer
Therefore, the limit is

\[ \boxed{3}. \]