Let f(x) be defined for all `x in R` such that `lim_(xrarr0) [f(x)+log(1-(1)/(e^(f(x))))-log(f(x))]=0`. Then f(0) is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to evaluate the limit given in the question:

\[ \lim_{x \to 0} \left[ f(x) + \log\left(1 - \frac{1}{e^{f(x)}}\right) - \log(f(x)) \right] = 0 \]

### Step 1: Simplify the expression inside the limit

We can rewrite the limit expression as:

\[ \lim_{x \to 0} \left[ f(x) + \log\left(1 - \frac{1}{e^{f(x)}}\right) - \log(f(x)) \right] \]

This can be simplified to:

\[ \lim_{x \to 0} \left[ f(x) + \log\left(\frac{1 - \frac{1}{e^{f(x)}}}{f(x)}\right) \right] = 0 \]

### Step 2: Analyze the logarithmic term

As \( x \to 0 \), we need to analyze the behavior of \( f(x) \). Assuming \( f(x) \to L \) as \( x \to 0 \), we can substitute \( L \) into the limit:

\[ \lim_{x \to 0} \left[ L + \log\left(\frac{1 - \frac{1}{e^{L}}}{L}\right) \right] = 0 \]

### Step 3: Set the limit to zero

For the limit to equal zero, we have:

\[ L + \log\left(\frac{1 - \frac{1}{e^{L}}}{L}\right) = 0 \]

This implies:

\[ \log\left(\frac{1 - \frac{1}{e^{L}}}{L}\right) = -L \]

### Step 4: Exponentiate both sides

Exponentiating both sides gives:

\[ \frac{1 - \frac{1}{e^{L}}}{L} = e^{-L} \]

### Step 5: Rearranging the equation

Rearranging the equation results in:

\[ 1 - \frac{1}{e^{L}} = L e^{-L} \]

### Step 6: Solve for \( L \)

Multiplying through by \( e^{L} \) gives:

\[ e^{L} - 1 = L \]

This is a transcendental equation. We can analyze it for potential solutions.

### Step 7: Check for specific values

By inspection, we can check \( L = 0 \):

\[ e^{0} - 1 = 0 \]

This holds true. Thus, we find \( L = 0 \).

### Conclusion

Since \( L = 0 \), we conclude that:

\[ f(0) = 0 \]