Let `f : R to R ` be a differentiable function satisfying `f'(3) + f'(2) = 0 `, Then ` lim_(x to 0) ((1+f(3+x)-f(3))/(1+f(2-x)-f(2)))^(1/x) ` is equal to

Options

A. e
B. `e^(-1)`
C. `e^(2)`
D. 1

Explanation

To solve the limit problem, we will follow these steps:

**Step 1: Rewrite the limit expression.**
We need to evaluate:
\[ \lim_{x \to 0} \left( \frac{1 + f(3+x) - f(3)}{1 + f(2-x) - f(2)} \right)^{\frac{1}{x}} \]

**Step 2: Simplify the expression inside the limit.**
As \( x \to 0 \), both the numerator and denominator approach 1, leading to an indeterminate form of \( \frac{0}{0} \). We can rewrite the limit using the exponential function:
\[ \lim_{x \to 0} \left( \frac{1 + f(3+x) - f(3)}{1 + f(2-x) - f(2)} \right)^{\frac{1}{x}} = e^{\lim_{x \to 0} \frac{f(3+x) - f(3) - (f(2-x) - f(2))}{x}} \]

**Step 3: Apply L'Hôpital's Rule.**
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{f(3+x) - f(3) - (f(2-x) - f(2))}{x} = \lim_{x \to 0} \frac{f'(3+x) + f'(2-x)}{1} \]

**Step 4: Evaluate the derivatives.**
From the problem, we know that:
\[ f'(3) + f'(2) = 0 \]
Thus, as \( x \to 0 \):
\[ f'(3+x) \to f'(3) \quad \text{and} \quad f'(2-x) \to f'(2) \]
So,
\[ \lim_{x \to 0} \left( f'(3+x) + f'(2-x) \right) = f'(3) + f'(2) = 0 \]

**Step 5: Substitute back into the limit.**
Now we can substitute this back into our limit:
\[ \lim_{x \to 0} \frac{f(3+x) - f(3) - (f(2-x) - f(2))}{x} = 0 \]

**Step 6: Final evaluation.**
Thus, we have:
\[ e^{\lim_{x \to 0} \frac{f(3+x) - f(3) - (f(2-x) - f(2))}{x}} = e^0 = 1 \]

Therefore, the final result is:
\[ \boxed{1} \]