If `lim_(xto0) [1+x+(f(x))/(x)]^(1//x)=e^(3)`, then the value of `ln(lim_(xto0) [1+(f(x))/(x)]^(1//x))` is _________.

Explanation

To solve the problem step by step, we start with the given limit:

\[ \lim_{x \to 0} \left[ 1 + x + \frac{f(x)}{x} \right]^{\frac{1}{x}} = e^3 \]

### Step 1: Analyze the limit
We notice that as \( x \to 0 \), the expression \( 1 + x + \frac{f(x)}{x} \) approaches \( 1 + 0 + \frac{f(0)}{0} \), which is an indeterminate form of \( 1^{\infty} \).

### Step 2: Rewrite the limit
Using the property of limits, we can express this limit in a more manageable form:

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

### Step 3: Simplify the expression
We can rewrite the limit as:

\[ \lim_{x \to 0} \left( \frac{f(x)}{x} + x \right) = 3 \]

This implies:

\[ \lim_{x \to 0} \frac{f(x)}{x} = 3 - \lim_{x \to 0} x = 3 \]

### Step 4: Find the limit of \( \frac{f(x)}{x} \)
From the above, we can conclude that:

\[ \lim_{x \to 0} \frac{f(x)}{x} = 3 \]

### Step 5: Find \( \lim_{x \to 0} \left[ 1 + \frac{f(x)}{x} \right]^{\frac{1}{x}} \)
Now we need to find:

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

This can also be rewritten as:

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

### Step 6: Substitute the limit
Using the previous result:

\[ \lim_{x \to 0} \frac{f(x)}{x} = 3 \implies \lim_{x \to 0} \left( \frac{f(x)}{x} - 1 \right) = 2 \]

Thus, we have:

\[ \lim_{x \to 0} \left[ 1 + \frac{f(x)}{x} \right]^{\frac{1}{x}} = e^2 \]

### Step 7: Find \( \ln \left( \lim_{x \to 0} \left[ 1 + \frac{f(x)}{x} \right]^{\frac{1}{x}} \right) \)
Now we need to find:

\[ \ln \left( \lim_{x \to 0} \left[ 1 + \frac{f(x)}{x} \right]^{\frac{1}{x}} \right) = \ln(e^2) = 2 \]

### Final Answer
Thus, the value of \( \ln \left( \lim_{x \to 0} \left[ 1 + \frac{f(x)}{x} \right]^{\frac{1}{x}} \right) \) is:

\[ \boxed{2} \]