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The given mathematical expression is:

\[
\lim_{x \to 0} \frac{\sin^3(2x)}{(x + 1) - \sqrt{10 + 2x}}
\]

### Steps to solve:

1. **Evaluate the limit**: The first step is to substitute \(x = 0\) into the expression to check if the limit exists.

   Substituting \(x = 0\):
   \[
   \frac{\sin^3(2(0))}{(0 + 1) - \sqrt{10 + 2(0)}} = \frac{0^3}{1 - \sqrt{10}} = 0
   \]
   The form is \(\frac{0}{0}\), which is indeterminate. Therefore, we can apply **L'Hôpital's Rule**.

2. **L'Hôpital's Rule**: Differentiate the numerator and denominator separately with respect to \(x\).

   - **Numerator**: The numerator is \(\sin^3(2x)\). Using the chain rule:
     \[
     \frac{d}{dx} \left(\sin^3(2x)\right) = 3\sin^2(2x) \cdot \frac{d}{dx} (\sin(2x)) = 3\sin^2(2x) \cdot 2\cos(2x) = 6\sin^2(2x)\cos(2x)
     \]

   - **Denominator**: The denominator is \((x + 1) - \sqrt{10 + 2x}\). Differentiating term by term:
     \[
     \frac{d}{dx} \left((x + 1) - \sqrt{10 + 2x}\right) = 1 - \frac{1}{2\sqrt{10 + 2x}} \cdot 2 = 1 - \frac{1}{\sqrt{10 + 2x}}
     \]

3. **Recompute the limit**: Now, apply the limit \(x \to 0\) to the differentiated expressions:

   - For the numerator:
     \[
     6\sin^2(2(0))\cos(2(0)) = 6 \cdot 0^2 \cdot 1 = 0
     \]
   - For the denominator:
     \[
     1 - \frac{1}{\sqrt{10 + 2(0)}} = 1 - \frac{1}{\sqrt{10}} = 1 - \frac{1}{\sqrt{10}}
     \]

Thus, the limit simplifies to:

\[
\lim_{x \to 0} \frac{0}{1 - \frac{1}{\sqrt{10}}} = 0
\]

### Final Answer:
The limit is \(0\).

Would you like a detailed explanation of any step or further clarification?

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Here are 5 related questions for practice:

1. What is L'Hôpital's Rule and when do we use it?
2. How do you differentiate a composite function like \(\sin^3(2x)\)?
3. What are common indeterminate forms that suggest using L'Hôpital's Rule?
4. How would this limit change if the sine function had a higher power?
5. How does the square root function affect the differentiation of the denominator?

**Tip**: Always check for indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying L'Hôpital's Rule.

Evaluate the limit: lim_{x→0} (sin^3(2x)) / ((x+1) - √(10+2x))

This problem involves evaluating a limit using L'Hôpital's Rule. The limit takes the indeterminate form 0/0, making it necessary to apply calculus techniques like differentiation. Suitable for advanced high school students and university-level learners.

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