Let `f:RrarrR` be such that `f(a)=1, f(a)=2`. Then `lim_(x to 0)((f^(2)(a+x))/(f(a)))^(1//x)` is

Options

A. `e^(2)`
B. `e^(4)`
C. `e^(-4)`
D. `1//e`

Explanation

To solve the limit problem step-by-step, we start with the given function and the limit we need to evaluate:

**Given:**
- \( f(a) = 1 \)
- \( f'(a) = 2 \)

**We need to evaluate:**
\[ \lim_{x \to 0} \left( \frac{f^2(a+x)}{f(a)} \right)^{\frac{1}{x}} \]

### Step 1: Substitute the known values
Since \( f(a) = 1 \), we can simplify the expression:
\[ \lim_{x \to 0} \left( \frac{f^2(a+x)}{1} \right)^{\frac{1}{x}} = \lim_{x \to 0} \left( f^2(a+x) \right)^{\frac{1}{x}} \]

### Step 2: Rewrite the limit
This limit can be rewritten using the exponential function:
\[ \lim_{x \to 0} \left( f^2(a+x) \right)^{\frac{1}{x}} = e^{\lim_{x \to 0} \frac{1}{x} \ln(f^2(a+x))} \]

### Step 3: Simplify the logarithm
Using properties of logarithms:
\[ \ln(f^2(a+x)) = 2 \ln(f(a+x)) \]
Thus, we have:
\[ \lim_{x \to 0} \frac{1}{x} \ln(f^2(a+x)) = \lim_{x \to 0} \frac{2 \ln(f(a+x))}{x} \]

### Step 4: Apply L'Hôpital's Rule
As \( x \to 0 \), \( f(a+x) \to f(a) = 1 \), which means \( \ln(f(a+x)) \to \ln(1) = 0 \). Therefore, we have a \( \frac{0}{0} \) form, and we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{2 \ln(f(a+x))}{x} = 2 \lim_{x \to 0} \frac{f'(a+x)}{f(a+x)} \]

### Step 5: Evaluate the limit
As \( x \to 0 \), \( f'(a+x) \to f'(a) = 2 \) and \( f(a+x) \to f(a) = 1 \):
\[ \lim_{x \to 0} \frac{f'(a+x)}{f(a+x)} = \frac{2}{1} = 2 \]

### Step 6: Substitute back into the limit
Thus:
\[ \lim_{x \to 0} \frac{2 \ln(f(a+x))}{x} = 2 \cdot 2 = 4 \]

### Step 7: Final result
Now substituting back into the exponential limit:
\[ e^{\lim_{x \to 0} \frac{2 \ln(f(a+x))}{x}} = e^4 \]

Therefore, the final answer is:
\[ \lim_{x \to 0} \left( \frac{f^2(a+x)}{f(a)} \right)^{\frac{1}{x}} = e^4 \]