Value of limit `lim_(xrarr0^(+))x^(x^(x)).ln(x)` is

Options

A. `1`
B. `0`
C. Does not exists
D. `2`

Explanation

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

### Step 1: Analyze \( x^{x^{x}} \)
First, we need to evaluate \( x^{x^{x}} \) as \( x \) approaches \( 0^+ \).

### Step 2: Evaluate \( x^{x} \)
We know that:
\[ x^{x} = e^{x \ln x} \]
To find \( \lim_{x \to 0^+} x^{x} \), we need to evaluate \( \lim_{x \to 0^+} x \ln x \).

### Step 3: Find \( \lim_{x \to 0^+} x \ln x \)
Using L'Hôpital's Rule, we rewrite:
\[ \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} \]
This is an indeterminate form \( \frac{-\infty}{\infty} \).

### Step 4: Apply L'Hôpital's Rule
Differentiating the numerator and denominator:
- Derivative of \( \ln x \) is \( \frac{1}{x} \)
- Derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \)

Thus, we have:
\[ \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0 \]
So, \( \lim_{x \to 0^+} x \ln x = 0 \).

### Step 5: Conclude \( \lim_{x \to 0^+} x^{x} \)
Since \( \lim_{x \to 0^+} x \ln x = 0 \), we have:
\[ \lim_{x \to 0^+} x^{x} = e^0 = 1 \]

### Step 6: Evaluate \( x^{x^{x}} \)
Now, substituting back, we find:
\[ x^{x^{x}} = x^{1} = x \]

### Step 7: Combine with \( \ln(x) \)
Now we evaluate:
\[ \lim_{x \to 0^+} x \ln x \]
We already established this limit as \( 0 \).

### Step 8: Final Limit
Thus, we conclude:
\[ \lim_{x \to 0^+} x^{x^{x}} \ln x = \lim_{x \to 0^+} x \ln x = 0 \]

### Final Answer
The value of the limit is:
\[ \boxed{0} \]