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The problem provided is:

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

We are asked to evaluate the limit of \( x^{\sin(x)} \) as \( x \) approaches 0 from the positive side (\( x \to 0^+ \)).

### Step 1: Understanding the behavior of the function
When \( x \to 0^+ \):
- \( \sin(x) \to 0 \) because \( \sin(0) = 0 \).
- \( x^{\sin(x)} = e^{\ln(x^{\sin(x)})} = e^{\sin(x) \ln(x)} \).

We will now analyze the exponent \( \sin(x) \ln(x) \) as \( x \to 0^+ \).

### Step 2: Behavior of \( \sin(x) \ln(x) \)
As \( x \to 0^+ \):
- \( \ln(x) \to -\infty \).
- \( \sin(x) \to 0 \).

The product \( \sin(x) \ln(x) \) is the key part to analyze. We use the approximation \( \sin(x) \approx x \) for small \( x \) (since \( \sin(x) \) behaves linearly near 0), so we approximate:

\[
\sin(x) \ln(x) \approx x \ln(x).
\]

Now, we evaluate the limit of \( x \ln(x) \) as \( x \to 0^+ \). This is a standard limit that evaluates to 0:

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

### Step 3: Conclusion
Since \( \sin(x) \ln(x) \to 0 \), we conclude:

\[
x^{\sin(x)} = e^{\sin(x) \ln(x)} \to e^0 = 1.
\]

Thus, the limit is:

\[
\lim_{x \to 0^+} x^{\sin(x)} = 1.
\]

Would you like further details or clarifications on any step?

Here are 5 related questions to explore:
1. How does the limit \( \lim_{x \to 0^+} x \ln(x) = 0 \) work?
2. What is L'Hôpital's Rule and how could it be used here?
3. How do we derive approximations like \( \sin(x) \approx x \) for small \( x \)?
4. Can you explain the general behavior of \( x^a \) when \( a \to 0 \)?
5. How does the logarithmic transformation help in solving exponential limits?

**Tip:** When dealing with limits that involve indeterminate forms like \( 0^0 \), logarithmic transformations are often helpful!

Evaluate the limit: lim (x → 0+) x^(sin(x))

Learn how to solve the limit lim (x → 0+) x^(sin(x)). This problem covers key calculus concepts including exponential functions, logarithmic transformations, and the small angle approximation sin(x) ≈ x. Suitable for university-level calculus students.

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