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 analyze 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: Rewrite the limit expression
We can rewrite the logarithmic terms using the property of logarithms:

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

Thus, the limit can be expressed as:

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

### Step 2: Analyze the limit as \( x \to 0 \)
As \( x \to 0 \), we need to consider the behavior of \( f(x) \). Let's denote \( f(0) = L \). The limit becomes:

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

### Step 3: Substitute \( f(0) \) into the limit
If we assume that \( f(0) = L \), we can substitute \( f(x) \) with \( L \) in the limit:

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

### Step 4: Solve for \( L \)
Rearranging gives us:

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

Exponentiating both sides gives:

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

This can be rewritten as:

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

### Step 5: Analyze the equation
Rearranging gives us:

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

### Step 6: Consider the case \( L = 0 \)
If we assume \( L = 0 \):

\[ 1 = 0 \cdot e^{0} + \frac{1}{e^{0}} = 1 \]

This holds true, suggesting that \( f(0) = 0 \) is a valid solution.

### Conclusion
Thus, we conclude that:

\[ f(0) = 0 \]