To evaluate the limit $\lim_{x \to 0^+} x^{\sin x}$, let's proceed step-by-step:

Step 1: Rewrite $x^{\sin x}$

The term $x^{\sin x}$ can be rewritten using the exponential and logarithmic identities: $x^{\sin x} = e^{\sin x \cdot \ln x}.$ Thus, the problem reduces to finding: $\lim_{x \to 0^+} e^{\sin x \cdot \ln x}.$

Step 2: Analyze the exponent $\sin x \cdot \ln x$

We now focus on evaluating: $\lim_{x \to 0^+} \sin x \cdot \ln x.$

1. As $x \to 0^+$, $\sin x \to 0$.
2. $\ln x \to -\infty$ as $x \to 0^+$.
3. The product $\sin x \cdot \ln x$ involves the interplay between a term approaching 0 ($\sin x$) and a term approaching $-\infty$ ($\ln x$).

Step 3: Approximation for $\sin x$ near 0

For small $x$, $\sin x \approx x$. Substituting this approximation: $\sin x \cdot \ln x \approx x \cdot \ln x.$

Now, analyze $\lim_{x \to 0^+} x \cdot \ln x$:

• Let $y = x \cdot \ln x$. Rewrite it as: $y = \frac{\ln x}{1/x}.$
• This is a $\frac{-\infty}{\infty}$ indeterminate form, so we apply L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(1/x)} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0.$

Thus, $\lim_{x \to 0^+} x \cdot \ln x = 0$, and consequently: $\lim_{x \to 0^+} \sin x \cdot \ln x = 0.$

Step 4: Exponentiation

Using the result from Step 3: $\lim_{x \to 0^+} e^{\sin x \cdot \ln x} = e^0 = 1.$

Final Answer:

$\lim_{x \to 0^+} x^{\sin x} = 1.$

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Would you like further clarification or a visual explanation? Here are some related questions:

1. How does $x \ln x$ behave as $x \to 0^+$?
2. Why is $\sin x \approx x$ a valid approximation for small $x$?
3. How does L'Hôpital's Rule apply in indeterminate forms like $-\infty \times 0$?
4. Can $x^{\sin x}$ ever exceed 1 near $x = 0$?
5. What are the implications of this limit in calculus or real-world contexts?

Tip: When dealing with limits involving exponentials and logarithms, rewriting the expression in terms of $e^u$ often simplifies the problem.