Slove `lim_(xto0)((1+x)^(1//x)-e)/x`

Answer

To solve the limit \( \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} \), we can follow these steps:

### Step 1: Recognize the form of the limit
First, we need to evaluate the expression \( (1+x)^{\frac{1}{x}} \) as \( x \) approaches 0. We know that:
\[ \lim_{x \to 0} (1+x)^{\frac{1}{x}} = e \]
This means that as \( x \to 0 \), \( (1+x)^{\frac{1}{x}} \) approaches \( e \).

### Step 2: Substitute into the limit
Substituting this into our limit gives us:
\[ \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} \]
This is an indeterminate form of type \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This involves differentiating the numerator and the denominator.

Let:
\[ f(x) = (1+x)^{\frac{1}{x}} - e \]
We need to differentiate \( f(x) \) and the denominator \( g(x) = x \).

### Step 4: Differentiate the numerator
To differentiate \( f(x) \), we first rewrite \( (1+x)^{\frac{1}{x}} \) using logarithms:
\[ y = (1+x)^{\frac{1}{x}} \implies \ln y = \frac{1}{x} \ln(1+x) \]
Differentiating both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\ln(1+x)}{x} \right) \]
Using the quotient rule:
\[ \frac{d}{dx} \left( \frac{\ln(1+x)}{x} \right) = \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} \]
Thus,
\[ \frac{dy}{dx} = y \cdot \left( \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} \right) \]
Substituting back \( y = (1+x)^{\frac{1}{x}} \):
\[ \frac{dy}{dx} = (1+x)^{\frac{1}{x}} \cdot \left( \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} \right) \]

### Step 5: Evaluate the limit using L'Hôpital's Rule
Now we apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} \cdot \left( \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} \right)}{1} \]
As \( x \to 0 \), \( (1+x)^{\frac{1}{x}} \to e \) and we need to evaluate the limit of the remaining expression.

### Step 6: Simplify the limit
We can simplify the limit:
\[ \lim_{x \to 0} \left( \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} \right) \]
This again gives an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule again.

### Step 7: Final evaluation
After applying L'Hôpital's Rule again and simplifying, we find that the limit evaluates to:
\[ \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} = \frac{1}{2} \]

### Final Answer
Thus, the final answer is:
\[ \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} = \frac{1}{2} \]