`lim_(x->0) ((1+x)^(1/x)-e)/x` is equal to

Options

A. e/2
B. e
C. #REF!
D. #REF!

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x}, \]

we can follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \), we know that \( (1+x)^{\frac{1}{x}} \) approaches \( e \). Therefore, we have:

\[ (1+x)^{\frac{1}{x}} \to e \quad \text{and} \quad e \to e. \]

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit is in the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator.

Let \( f(x) = (1+x)^{\frac{1}{x}} - e \) and \( g(x) = x \). We need to find \( f'(x) \) and \( g'(x) \).

### Step 3: Differentiate the numerator
To differentiate \( f(x) = (1+x)^{\frac{1}{x}} - e \), we first differentiate \( (1+x)^{\frac{1}{x}} \):

Using the logarithmic differentiation:

\[ y = (1+x)^{\frac{1}{x}} \implies \ln y = \frac{1}{x} \ln(1+x). \]

Differentiating both sides with respect to \( x \):

\[ \frac{1}{y} \frac{dy}{dx} = -\frac{1}{x^2} \ln(1+x) + \frac{1}{x(1+x)}. \]

Thus,

\[ \frac{dy}{dx} = y \left( -\frac{1}{x^2} \ln(1+x) + \frac{1}{x(1+x)} \right). \]

Substituting back \( y = (1+x)^{\frac{1}{x}} \):

\[ f'(x) = (1+x)^{\frac{1}{x}} \left( -\frac{1}{x^2} \ln(1+x) + \frac{1}{x(1+x)} \right). \]

### Step 4: Differentiate the denominator
The derivative of \( g(x) = x \) is simply:

\[ g'(x) = 1. \]

### Step 5: Apply L'Hôpital's Rule
Now we apply L'Hôpital's Rule:

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

### Step 6: Evaluate the limit
As \( x \to 0 \):

- \( (1+x)^{\frac{1}{x}} \to e \)
- \( \ln(1+x) \to x \) (using the Taylor expansion)
- Therefore, \( -\frac{1}{x^2} \ln(1+x) \to -\frac{x}{x^2} = -\frac{1}{x} \to -\infty \) and \( \frac{1}{x(1+x)} \to \frac{1}{x} \).

Thus, we can simplify the limit:

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

### Final Result
The limit evaluates to:

\[ \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} = -\frac{e}{2}. \]