To find the asymptotic equivalence of $\sin(\sqrt{1+x} - 1)$ as $x \to 0$, let's examine this expression closely and use Taylor expansions for small $x$.

1. Expand $\sqrt{1+x}$:

   For $x \to 0$, we can use the Taylor series for $\sqrt{1 + x}$: $\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^2}{8} + O(x^3).$ So, $\sqrt{1 + x} - 1 \approx \frac{x}{2}.$

2. Consider $\sin(\sqrt{1+x} - 1)$:

   Substitute $\sqrt{1+x} - 1 \approx \frac{x}{2}$ into $\sin(\sqrt{1+x} - 1)$: $\sin(\sqrt{1+x} - 1) \approx \sin\left(\frac{x}{2}\right).$

3. Expand $\sin\left(\frac{x}{2}\right)$:

   For $x \to 0$, we can expand $\sin\left(\frac{x}{2}\right)$ as follows: $\sin\left(\frac{x}{2}\right) \approx \frac{x}{2} - \frac{1}{6} \left(\frac{x}{2}\right)^3 + O(x^5).$ Simplifying, we get $\sin\left(\frac{x}{2}\right) \approx \frac{x}{2} - \frac{x^3}{48} + O(x^5).$

4. Conclusion:

   As $x \to 0$, $\sin(\sqrt{1+x} - 1) \sim \frac{x}{2}.$

Thus, the small equivalence of $\sin(\sqrt{1+x} - 1)$ as $x \to 0$ is approximately $\frac{x}{2}$.

Would you like more details on any part of this calculation, or have any questions?

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Here are 5 related questions you might find interesting:

1. How does this approach change if we consider $\cos(\sqrt{1+x} - 1)$ instead?
2. What happens if we look for the equivalence as $x \to \infty$ instead of $x \to 0$?
3. How can we apply Taylor expansions to other trigonometric functions in similar contexts?
4. How would this change if we were to analyze $\tan(\sqrt{1+x} - 1)$ instead?
5. How can we generalize this approach to other functions composed of radicals and trigonometric terms?

Tip: Remember that for small values of $x$, Taylor series expansions can simplify expressions significantly, providing approximate values that are often easier to work with.