To find the limit of $\lim_{x \to 0^+} (\sin x)^x$, we can approach the problem step by step.

Step 1: Express the limit in a more manageable form

We start by expressing the given limit:

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

Since directly evaluating the expression might be challenging, we can take the natural logarithm of the expression. This will allow us to simplify the exponentiation:

$y = (\sin x)^x$

Now take the natural logarithm of both sides:

$\ln y = \ln\left( (\sin x)^x \right) = x \ln(\sin x)$

Thus, we need to find the limit of $x \ln(\sin x)$ as $x \to 0^+$.

Step 2: Evaluate the limit of $x \ln(\sin x)$

We know that $\sin x \approx x$ when $x$ is close to 0, so we can approximate $\ln(\sin x)$ by $\ln(x)$ for small values of $x$. Therefore, the limit becomes:

$\lim_{x \to 0^+} x \ln(\sin x) \approx \lim_{x \to 0^+} x \ln(x)$

Step 3: Evaluate the limit of $x \ln(x)$

We can evaluate the limit of $x \ln(x)$ using L'Hopital's Rule. The limit is of the indeterminate form $0 \times (-\infty)$, so we rewrite it as a fraction:

$\lim_{x \to 0^+} x \ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{1/x}$

Now apply L'Hopital's Rule. Differentiate the numerator and denominator:

$\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$

Thus:

$\lim_{x \to 0^+} x \ln(x) = 0$

Step 4: Conclude the original limit

Since $x \ln(\sin x) \to 0$, we have:

$\lim_{x \to 0^+} \ln y = 0$

Exponentiating both sides:

$y = e^0 = 1$

Thus:

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

Final Answer:

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

Would you like further details or clarification on any part of the solution?

Here are some related questions you might find interesting:

1. How does L'Hopital's Rule apply to indeterminate forms?
2. What is the approximation $\sin x \approx x$ valid for?
3. How can we use logarithms to simplify limits involving exponents?
4. How do we handle limits involving natural logarithms and small values of $x$?
5. Can you compute limits of the form $0 \times (-\infty)$ without L'Hopital's Rule?

Tip: When dealing with limits involving exponentiation or small values of $x$, taking the logarithm often simplifies the expression and makes the limit easier to evaluate.