Evaluate`lim_(xto0^(+)) x^(x)"and "lim_(xto0^(+))x^(x^(x))`

Options

A. `1//2`
B. 2
C. 1
D. None of these

Explanation

To evaluate the limits \( \lim_{x \to 0^+} x^x \) and \( \lim_{x \to 0^+} x^{x^x} \), we will proceed step by step.

### Step 1: Evaluate \( \lim_{x \to 0^+} x^x \)

1. **Set \( y = x^x \)**: We start by letting \( y = x^x \).
2. **Take the logarithm**: To simplify the expression, we take the natural logarithm of both sides:
\[ \log y = \log(x^x) = x \log x \]
3. **Rewrite the limit**: We need to evaluate the limit as \( x \) approaches \( 0^+ \):
\[ \lim_{x \to 0^+} \log y = \lim_{x \to 0^+} x \log x \]
4. **Change the form**: Notice that \( x \log x \) approaches the form \( 0 \cdot (-\infty) \). We can rewrite it as:
\[ \lim_{x \to 0^+} x \log x = \lim_{x \to 0^+} \frac{\log x}{1/x} \]
5. **Apply L'Hôpital's Rule**: This is now in the form \( \frac{-\infty}{\infty} \), so we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0^+} \frac{\log x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0 \]
6. **Conclude for \( y \)**: Since \( \lim_{x \to 0^+} \log y = 0 \), we have:
\[ \log y \to 0 \implies y \to e^0 = 1 \]
Thus,
\[ \lim_{x \to 0^+} x^x = 1 \]

### Step 2: Evaluate \( \lim_{x \to 0^+} x^{x^x} \)

1. **Use the result from Step 1**: We know that \( \lim_{x \to 0^+} x^x = 1 \).
2. **Set up the limit**: Now we evaluate:
\[ \lim_{x \to 0^+} x^{x^x} = \lim_{x \to 0^+} x^{1} = \lim_{x \to 0^+} x = 0 \]

### Final Answers:
- \( \lim_{x \to 0^+} x^x = 1 \)
- \( \lim_{x \to 0^+} x^{x^x} = 0 \)