Let `f(x^m y^n)=mf(x)+nf(y)` for all `x , y in R^+` and for all `m ,n in Rdot` If `f^(prime)(x)` exists and has the value `e/x ,` then find `(lim)_(xvec0)(f(1+x))/x`

Explanation

To solve the problem step by step, we start with the given functional equation and the derivative information.

### Step 1: Analyze the functional equation
We have the functional equation:
\[ f(x^m y^n) = mf(x) + nf(y) \]
for all \( x, y \in \mathbb{R}^+ \) and for all \( m, n \in \mathbb{R} \).

### Step 2: Substitute specific values
Let's set \( x = 1 \) and \( y = 1 \), \( m = 1 \), and \( n = 1 \):
\[ f(1^1 \cdot 1^1) = 1 \cdot f(1) + 1 \cdot f(1) \]
This simplifies to:
\[ f(1) = 2f(1) \]
Subtracting \( f(1) \) from both sides gives:
\[ 0 = f(1) \]
Thus, we find:
\[ f(1) = 0 \]

### Step 3: Set up the limit
We need to find:
\[ \lim_{x \to 0} \frac{f(1+x)}{x} \]
Since \( f(1) = 0 \), substituting \( x = 0 \) gives us a \( \frac{0}{0} \) form.

### Step 4: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limits exist.

In our case:
- \( f(x) = f(1+x) \)
- \( g(x) = x \)

Differentiating both:
- The derivative of the numerator \( f(1+x) \) with respect to \( x \) is \( f'(1+x) \).
- The derivative of the denominator \( x \) is \( 1 \).

Thus, we have:
\[ \lim_{x \to 0} \frac{f(1+x)}{x} = \lim_{x \to 0} f'(1+x) \]

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

### Step 6: Use the given derivative information
We are given that:
\[ f'(x) = \frac{e}{x} \]
Thus,
\[ f'(1) = \frac{e}{1} = e \]

### Conclusion
The limit we are looking for is:
\[ \lim_{x \to 0} \frac{f(1+x)}{x} = f'(1) = e \]

Therefore, the final answer is:
\[ \boxed{e} \]

---