To evaluate the limit

$\lim_{x \to 0^+} x^{\sin x},$

we'll proceed as follows:

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Step 1: Rewrite the expression

The term $x^{\sin x}$ can be rewritten using the exponential function: $x^{\sin x} = e^{\sin x \cdot \ln x}.$

So, the limit becomes: $\lim_{x \to 0^+} x^{\sin x} = \lim_{x \to 0^+} e^{\sin x \cdot \ln x}.$

Since the exponential function $e^y$ is continuous, this reduces to: $e^{\lim_{x \to 0^+} (\sin x \cdot \ln x)}.$

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

As $x \to 0^+$:

• $\sin x \to 0$, and
• $\ln x \to -\infty$.

Thus, the product $\sin x \cdot \ln x$ has an indeterminate form $0 \cdot (-\infty)$. To resolve this, rewrite it: $\sin x \cdot \ln x = \ln x \cdot \sin x = \frac{\ln x}{1/\sin x}.$

Now, as $x \to 0^+$, $\sin x \to 0$, so $1/\sin x \to \infty$. This results in the new indeterminate form $\frac{-\infty}{\infty}$, which allows us to apply L'Hôpital's Rule.

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Step 3: Apply L’Hôpital’s Rule

We evaluate: $\lim_{x \to 0^+} \frac{\ln x}{1/\sin x}.$

Differentiating the numerator and denominator:

• The derivative of $\ln x$ is $\frac{1}{x}$,
• The derivative of $1/\sin x$ is $-\frac{\cos x}{\sin^2 x}$.

Thus: $\lim_{x \to 0^+} \frac{\ln x}{1/\sin x} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{\cos x}{\sin^2 x}} = \lim_{x \to 0^+} \frac{-\sin^2 x}{x \cos x}.$

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Step 4: Simplify the new expression

As $x \to 0^+$:

• $\sin^2 x \to 0$,
• $x \to 0$,
• $\cos x \to 1$.

The expression simplifies to: $\frac{-\sin^2 x}{x \cos x} = \frac{-\sin^2 x}{x}.$

We now focus on $\frac{\sin^2 x}{x}$. Using the small-angle approximation $\sin x \sim x$ as $x \to 0$: $\frac{\sin^2 x}{x} \sim \frac{x^2}{x} = x.$

Thus: $\frac{-\sin^2 x}{x} \to 0 \quad \text{as } x \to 0^+.$

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Step 5: Substitute back

We found that: $\lim_{x \to 0^+} \sin x \cdot \ln x = 0.$

Returning to the original limit: $\lim_{x \to 0^+} x^{\sin x} = e^0 = 1.$

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Final Answer:

$\boxed{1}$

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Would you like more details or have any questions? Here are some related concepts for exploration:

1. Why does $x^{\sin x}$ need to be rewritten as $e^{\sin x \ln x}$?
2. How does the small-angle approximation $\sin x \sim x$ work for limits?
3. What are common indeterminate forms resolved by L'Hôpital's Rule?
4. Why is the exponential function $e^y$ continuous and critical in limit transformations?
5. Can you extend this problem to other forms like $\lim_{x \to 0^+} x^{\cos x}$?

Tip: Always rewrite power functions like $x^f$ in terms of exponentials for easier limit evaluation!