The value of `lim_(xrarr0^(+)){x^(x)+x^((x^(x)))}` is equal to

Options

A. 0
B. 1
C. 2
D. `(1)/(2)`

Answer

To solve the limit \( \lim_{x \to 0^+} \left( x^x + x^{(x^x)} \right) \), we can break it down into two parts: \( x^x \) and \( x^{(x^x)} \).

### Step 1: Evaluate \( x^x \)

We start with the first part, \( x^x \). We can rewrite it using the exponential function:

\[ x^x = e^{x \ln x} \]

As \( x \to 0^+ \), we need to evaluate \( x \ln x \).

### Step 2: Analyze \( x \ln x \)

We know that \( \ln x \to -\infty \) as \( x \to 0^+ \). Therefore, we analyze the product \( x \ln x \):

\[ \lim_{x \to 0^+} x \ln x \]

This is an indeterminate form \( 0 \cdot (-\infty) \). We can rewrite it as:

\[ \lim_{x \to 0^+} \frac{\ln x}{1/x} \]

### Step 3: Apply L'Hôpital's Rule

This limit is now in the form \( \frac{-\infty}{\infty} \), so we apply L'Hôpital's Rule:

\[ \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0 \]

Thus, we find:

\[ \lim_{x \to 0^+} x \ln x = 0 \]

### Step 4: Find \( x^x \)

Now substituting back, we have:

\[ \lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x \ln x} = e^0 = 1 \]

### Step 5: Evaluate \( x^{(x^x)} \)

Next, we consider the second part, \( x^{(x^x)} \). Since we found that \( x^x \to 1 \) as \( x \to 0^+ \), we can rewrite it as:

\[ x^{(x^x)} = x^1 = x \]

### Step 6: Find the limit of \( x^{(x^x)} \)

Now we evaluate:

\[ \lim_{x \to 0^+} x^{(x^x)} = \lim_{x \to 0^+} x = 0 \]

### Step 7: Combine the results

Now we combine the results from both parts:

\[ \lim_{x \to 0^+} \left( x^x + x^{(x^x)} \right) = \lim_{x \to 0^+} x^x + \lim_{x \to 0^+} x^{(x^x)} = 1 + 0 = 1 \]

### Final Answer

Thus, the value of the limit is:

\[ \boxed{1} \]