What is wrong with the following application of L'Hopital's rule ?
`lim_(x to 0) ((1+x)^(1//x) -e)/x`

Answer

To analyze the limit and identify what is wrong with the application of L'Hôpital's rule, we start with the given limit:

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

### Step 1: Identify the form of the limit

First, we need to evaluate the limit as \( x \) approaches 0. We can substitute \( x = 0 \) directly into the expression:

\[ (1+0)^{\frac{1}{0}} - e = 1^{\infty} - e \]

This is an indeterminate form of type \( 0/0 \) since \( 1^{\infty} \) is not well-defined. Thus, we have:

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

### Step 2: Apply L'Hôpital's Rule

Since we have the indeterminate form \( 0/0 \), we can apply L'Hôpital's rule, which states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in \( 0/0 \) or \( \infty/\infty \), then:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

Here, let:

- \( f(x) = (1+x)^{\frac{1}{x}} - e \)
- \( g(x) = x \)

We need to differentiate both \( f(x) \) and \( g(x) \).

### Step 3: Differentiate the numerator

To differentiate \( f(x) \), we can use the chain rule and logarithmic differentiation:

Let \( y = (1+x)^{\frac{1}{x}} \). Taking the natural logarithm on both sides gives:

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

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 \( y \) back in:

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

### 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 can apply L'Hôpital's rule:

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

### Step 6: Evaluate the limit

As \( x \to 0 \), \( (1+x)^{\frac{1}{x}} \to e \) and the other terms need to be evaluated carefully. However, the differentiation process can lead to further complications, and we may need to apply L'Hôpital's rule multiple times.

### Conclusion: What is wrong?

The application of L'Hôpital's rule is valid, but the complexity of the derivatives and the potential for further indeterminate forms must be handled with care. The original function \( (1+x)^{\frac{1}{x}} \) approaches \( e \) as \( x \to 0 \), and thus the limit simplifies to evaluating derivatives correctly.