To evaluate $\lim_{x \to 0^+} (1 - \cos x)^{\sin x}$, let us break it into manageable parts.

Step-by-step solution:

1. Rewrite the expression: $y = (1 - \cos x)^{\sin x}$

   Taking the natural logarithm on both sides: $\ln y = \sin x \cdot \ln(1 - \cos x)$

   The limit now becomes: $\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} \sin x \cdot \ln(1 - \cos x)$

2. Analyze $\ln(1 - \cos x)$: As $x \to 0^+$, $(1 - \cos x)$ approaches $0$. Using the approximation for small $x$: $1 - \cos x \approx \frac{x^2}{2}$ Thus: $\ln(1 - \cos x) \approx \ln\left(\frac{x^2}{2}\right) = \ln(x^2) - \ln 2 = 2 \ln x - \ln 2$

3. Substitute back: Substituting this into $\ln y$, we have: $\ln y \approx \sin x \cdot \left(2 \ln x - \ln 2\right)$

4. Analyze $\sin x$: As $x \to 0^+$, $\sin x \approx x$. Substituting: $\ln y \approx x \cdot \left(2 \ln x - \ln 2\right) = 2x \ln x - x \ln 2$

5. Evaluate each term:

   • For $2x \ln x$: $\lim_{x \to 0^+} 2x \ln x = \lim_{x \to 0^+} \frac{2 \ln x}{1/x}$ Using L’Hôpital’s Rule: $= \lim_{x \to 0^+} \frac{2/x}{-1/x^2} = \lim_{x \to 0^+} -2x = 0$

   • For $-x \ln 2$: $\lim_{x \to 0^+} -x \ln 2 = 0$

   Thus, $\ln y \to 0$.

6. Exponentiate to get the result: $y = e^{\ln y} \to e^0 = 1$

Final Answer:

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

---

Would you like a deeper explanation of any step?

Here are five related questions to expand your understanding:

1. How is L’Hôpital’s Rule applied in limits involving indeterminate forms?
2. Why is $(1 - \cos x) \approx \frac{x^2}{2}$ a good approximation for small $x$?
3. Can you explain why $\ln(x)$ dominates as $x \to 0^+$?
4. What happens if the base in this limit were different, e.g., $(1 - \cos x)^x$?
5. How does Taylor expansion help in solving limits involving trigonometric functions?

Tip: For exponential and logarithmic forms in limits, consider logarithmic properties to simplify calculations.